Hc Verma I for Class 11 Science Physics Chapter 17 - Light Waves

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Answer:

White light is a composition of seven colours. These are violet, indigo, blue, green, yellow, orange and red, collectively known as VIBGYOR. Spectrum of white light consists of sever colour bands. Each band consists of some range of wavelengths or frequencies.
For orange colour : (590 nm to 620 nm)
For red colour: (620 nm to 780 nm)

So, the colour of 620 nm and 780 nm lights may be different. But the colour of 620 nm light and 621 nm light is same.

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Colour of light will depend only on the frequency of light and not on the wavelength of the light. So, light will appear red to an observer under water.

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The width of the central band is inversely proportional to the slit width. So, as the width of the slit is increased, the central band will become less wider and further bands will start merging in them. Hence, diffraction effects will be visible less clearly.

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Light waves have the property of travelling in a straight line, unlike sound waves. When we put a cardboard between the light source and our eyes, the light waves are obstructed by the cardboard and cannot reach our eyes, which doesn't happen when the cardboard is inserted between sound source and our ear.

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To receive TV signals transmitted from Delhi in Patna directly, one has to use antennas of great height, which will cost much. On the other hand, transmission of signals with the help of satellites requires only high frequency waves and can be done easily.

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Answer:

Young's double slit experiment can be performed with sound waves, as the sound waves also show interference pattern. To get a reasonable "fringe pattern", the separation of the slits should be of the order of the wavelength of the sound waves used.
In this experiment, the bright and dark fringes can be detected by measuring the intensity of sound using a microphone or any other detector.

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Answer:

Interference pattern can be studied with waves of unequal intensity.
$Ι = Ι_{1} + Ι_{2} + 2 \sqrt{[Ι_{1} \times Ι_{2}]} \cos (ϕ)$ ,
where $ϕ = phase difference$ .
In case of waves of unequal intensities, the contrast will not be clear, as the minima will not be completely dark.

Page No 379:

Answer:

The interference pattern can be produced by any two coherent waves moving in the same direction. It cannot be concluded from the interference phenomenon that light is a transverse wave, as sound waves that are longitudinal in nature also interfere.

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In order to get interference, the sources should be coherent, i.e. they should emit wave of the same frequency and a stable phase difference. Two candles that are placed close to each other are distinct and cannot be considered as coherent sources. Two independent sources cannot be coherent. So, two different laser sources will also not serve the purpose.

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Answer:

The fringe width in Young's double slit experiment depends on the separation of the slits.
$χ = \frac{λ D}{d}$ ,
where
$λ = wavelength χ = fringe width D = distance between slits and screen d = separation between slits$
On increasing d, fringe width decreases. If the separation is increased too much, the fringes will merge with each other and the fringe pattern won't be detectable.

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Answer:

The violet filter will allow only violet light to pass through it. Now, if the double slit experiment is performed with the white light and violet light, the fringe pattern will not be the same as obtained by just using white light as the source. To have interference pattern, the light waves entering from the slits should be monochromatic. So, in this case, the violet light will superimpose with only violet light (of wavelength 400 nm) in such a way that the bright bands will be of violet colour and the minima will be completely dark.

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Answer:

(c) both a particle and a wave phenomenon

Light shows photoelectric effect and Compton effect, which depicts its particle nature. It also shows interference and diffraction, which depicts the wave nature of light.

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(d) neither on elasticity nor on inertia

The speed of light in any medium depends on the refractive index of that medium, which is an intensive property. Hence, speed of light is not affected by the elasticity and inertia of the medium.

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(d) electric field

Light consists of mutually perpendicular electric and magnetic fields. So, the equation of a light wave is represented by its field vector.

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(d) Polarization

Reflection, interference and diffraction are the phenomena shown by both transverse waves and longitudinal waves. Polarization is the phenomenon shown only by transverse waves.

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(c) its wavelength decreases but frequency remains unchanged

Frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.

Decrease in the wavelength of light entering a medium of refractive index $μ$ is given by
$λ_{Μ} = \frac{λ}{μ}, w h e r e λ_{Μ} = w a v e l e n g t h i n m e d i u m λ = w a v e l e n g t h i n v a c u u m μ = r e f r a c t i v e i n d e x$

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(b) Frequency

Frequency of a light wave doesn't change on changing the medium of propagation of light.

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(a) $λ_{a} > λ_{f}$

An electromagnetic wave bends round the corners of an obstacle if the size of the obstacle is comparable to the wavelength of the wave. An AM wave has less frequency than an FM wave. So, an AM wave has a higher wavelength than an FM wave and it bends round the corners of a 1 m $\times$ 1m board.
λ

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(d) A laser

Among the given sources, laser is the best coherent source providing monochromatic light with constant phase difference.

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(d) $\cos^{- 1} (1 / \sqrt{3})$
On writing the given equation in the plane equation form lx + my + nz = p,
where l² + m² + n² = 1 and p>0, we get:

$\frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = \frac{c}{\sqrt{3}}$
If $θ$ is the angle between the normal and +X axis, then
$\cos θ = \frac{1}{\sqrt{3}} \Rightarrow θ = \cos^{- 1} (\frac{1}{\sqrt{3}})$

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(a) plane

Wave travelling from a distant source always has plane wavefront.

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(a) point source

Intensity of a point source obeys the inverse square law.
Intensity of light at distance r from the point source is given by
$I = S / (4 {πr}^{2})$ ,
where S is the source strength.

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(d) having a constant phase difference
For light waves emitted by two sources of light to remain coherent, the initial phase difference between waves should remain constant in time. If the phase difference changes continuously or randomly with time, then the sources are incoherent.

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(d) interference of light

Interference effect is produced by a thin film ( coating of a thin layer of a translucent material on a medium of different refractive index which allows light to pass through it))In the present case, oil floating on water forms a thin film on the surface of water, leading to the display of beautiful colours in daylight because of the interference of sunlight.

Page No 380:

Answer:

Given:
Range of wave length is
$400 nm < λ < 700 nm$
We know that frequency is given by $f = \frac{c}{λ}$ ,
$where c = speed of light = 3 \times 10^{8} m / s$
f is the frequency
λ is the wavelength
We can write wavelength as:
$\frac{1}{700 nm} < \frac{1}{λ} < \frac{1}{400 nm} \Rightarrow \frac{1}{7 \times 10^{- 7} m} < \frac{1}{λ} < \frac{1}{4 \times 10^{- 7} m} \frac{3 \times 10^{8}}{7 \times 10^{- 7}} Hz < \frac{c}{λ} < \frac{3 \times 10^{8}}{4 \times 10^{- 7}} Hz \Rightarrow 4.3 \times 10^{14} Hz < \frac{c}{λ} < 7.5 \times 10^{14} Hz \Rightarrow 4.3 \times 10^{14} Hz < f < 7.5 \times 10^{14} Hz$

Hence, frequency of the range of light that is visible to an average human being is $4.3 \times 10^{14} Hz to 7.5 \times 10^{14} Hz$ .

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Answer:

Given:
Wavelength of sodium light in air, $λ_{a} = 589 nm = 589 \times 10^{- 9} m$
Refractive index of water, μ_w= 1⋅33
We know that $f = \frac{c}{λ}$ ,
$where c = speed of light = 3 \times 10^{8} m / s$
          f = frequency
          λ = wavelength
(a) Frequency in air, $f_{air} = \frac{c}{λ_{a}}$
$f_{air} = \frac{3 \times 10^{8}}{589 \times 10^{- 9}} = 5.09 \times 10^{14} Hz$

(b)
Let wavelength of sodium light in water be $λ_{w}$ .
We know that
$\frac{μ_{a}}{μ_{w}} = \frac{λ_{ω}}{λ_{a}}$ ,
where μ_a is the refractive index of air which is equal to 1 and
           λ_w is the wavelength of sodium light in water.
$\Rightarrow \frac{1}{1.33} = \frac{λ_{ω}}{589 \times 10^{- 9}} \Rightarrow λ_{ω} = 443 nm$

(c) Frequency of light does not change when light travels from one medium to another.
$∴ f_{ω} = f_{a} = 5.09 \times 10^{14} Hz$

(d) Let the speed of sodium light in water be $ν_{ω}$
     and speed in air, v_a = c.
     Using $\frac{μ_{a}}{μ_{ω}} = \frac{ν_{ω}}{ν_{a}}$ , we get:
$ν_{ω} = \frac{μ_{a} c}{μ_{ω}} = \frac{3 \times 10^{8}}{(1.33)} = 2.25 \times 10^{8} m / s$

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Answer:

Given:
Refractive index of fused quartz for light of wavelength 400 nm is 1.472.
And refractive index of fused quartz for light of wavelength 760 nm is 1.452.
We known that refractive index of a material is given by
μ = $\frac{Speed of light in vacuum}{Speed of light in the material} = \frac{c}{v}$
Let speed of light for wavelength 400 nm in quartz be v₄₀₀.
So,
$1.472 = \frac{3 \times 10^{8}}{v_{400}} \Rightarrow v_{400} = 2.04 \times 10^{8} m / s$
Let speed of light of wavelength 760 nm in quartz be v₇₆₀.
Again, $\frac{1.452}{1} = \frac{3 \times 10^{8}}{ν_{760}}$
$\Rightarrow v_{760} = 2.07 \times 10^{8} m / s$

Page No 380:

Answer:

Given:
Speed of yellow light in liquid (v_L)= 2⋅4 × 10⁸ m s⁻¹
And speed of yellow light in air speed = v_a
Let μ_L be the refractive index of the liquid
Using, $μ_{L} = \frac{speed of light in vaccum}{velocity of light in the given medium} = \frac{c}{v_{L}}$
$μ_{L} = \frac{3 \times 10^{8}}{(2.4) \times 10^{8}} = 1.25$

Hence, the required refractive index is 1.25.

Page No 380:

Answer:

Given:
Separation between two narrow slits, d = 1 cm = 10⁻² m
Wavelength of the light, $λ = 5 \times 10^{- 7} m$
Distance of the screen, $D = 1 m$
(a)
    We know that separation between two consecutive maxima = fringe width (β).
That is, $β = \frac{λ D}{d}$     ...(i)
       $= \frac{5 \times 10^{- 7} \times 1}{10^{- 2}} m = 5 \times 10^{- 5} m = 0.05 mm$

(b)
Separation between two consecutive maxima = fringe width
∴ $β = 1 mm = 10^{- 3} m$
Let the separation between the sources be 'd'
Using equation (i), we get:
$d' = \frac{5 \times 10^{- 7} \times 1}{10^{- 3}} \Rightarrow d' = 5 \times 10^{- 4} m = 0.50 mm .$

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Answer:

(c) 9 : 4

Ratio of maximum intensity and minimum intensity is given by
$\frac{I_{m a x}}{I_{m i n}} = \frac{{(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}}{{(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}} = \frac{25}{1} \Rightarrow \sqrt{I_{1}} = 3 a n d \sqrt{I_{2}} = 2 \Rightarrow I_{1} = 9 a n d I_{2} = 4$
Then,
$\frac{I_{1}}{I_{2}} = \frac{9}{4}$

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Answer:

(b) I₀/4

Total intensity coming from the source is I₀which is present at the central maxima. In case of two slits, the intensity is getting distributed between the two slits and for a single slit, the amplitude of light coming from the slit is reduced to half which leads to 1/4th of intensity.

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(c) remain same

On the introduction of a transparent sheet in front of one of the slits, the fringe pattern will shift slightly but the width will remain the same.

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(a) the fringe width will decrease

As fringe width is proportional to the wavelength and wavelength of light is inversely proportional to the refractive index of the medium,
Here, $λ_{Μ} = λ / η λ_{Μ} = wavelength in medium λ = wavelength in vacuum η = refractive index of medium$
Hence, fringe width decreases when Young's double slit experiment is performed under water.

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Answer:

(a) in vacuum
(c) in a material medium

Light is an electromagnetic wave that can travel through vacuum or any optical medium.

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(b) Speed of light in water is smaller than its speed in vacuum.
(c) Light shows interference.

Snell's Law, which states that the speed of light reduces on moving from a rarer to a denser medium, can be concluded from Huygens' wave theory and interference of light waves is based on the wave properties of light.

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Answer:

(b) have zero average value
(c) are perpendicular to the direction of propagation of light
(d) are mutually perpendicular

Light is an electromagnetic wave that propagates through its electric and magnetic field vectors, which are mutually perpendicular to each other, as well as to the direction of propagation of light. The average value of both the fields is zero.

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Answer:

(c) find the new position of a wavefront
(d) explain Snell's Law

Huygen's wave theory explains the origin of points for the new wavefront proceeding successively. It also explains the variation in speed of light on moving from one medium to another, i.e. it proves Snell's Law.

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Answer:

Since the speed of light is a universal constant, v_A₌v_B₌ v_C=

3 \times 10^{8}

m/s.

ν_{B} = \frac{1}{2} (v_{A} + v_{C})

. This expression also implies that v_A₌v_B₌ v_C_.
1

νs=12

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Answer:

(a) v_A> v_B> v_C
(d) v_B= (v_A + v_C)/2

In any other medium, the speed of light is given by $v = c / η, where η is the refractive index of the medium$ and according to Doppler effect, for an observer moving towards the source ,speed of light appears to be more than the other two cases. On the other hand, it will be least when the observer is moving away from the source.

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Answer:

(a) x = c

The wave is travelling along the X-axis. So, it'll have planar wavefront perpendicular to the X-axis.

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(b) consecutive fringes will come closer

Fringe width, $β = λ D / d$ .
Wavelength of red light is greater than wavelength of violet light; so, the fringe width will reduce.

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(a) The central fringe will be white.
(b) There will not be a completely dark fringe.
(d) The fringe next to the central will be violet.

The superposition of all the colours at the central maxima gives the central band a white colour. As we go from the centre to corner, the fringe colour goes from violet to red. There will not be a completely dark fringe, as complete destructive interference does not take place.

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Answer:

(a) (i) and (ii)
(d) (iii) and (iv).

The waves are travelling with the same frequencies and varying by constant phase difference.

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Given:
Separation between consecutive dark fringes = fringe width (β) = 1 mm = 10⁻³ m
Distance between screen and slit (D) = 2.5 m
The separation between slits (d) = 1 mm = 10⁻³ m
Let the wavelength of the light used in experiment be λ.
We know that
$β = \frac{λ D}{d}$
$10^{- 3} m = \frac{2.5 \times λ}{10^{- 3}} \Rightarrow λ = \frac{1}{2.5} 10^{- 6} m = 4 \times 10^{- 7} m = 400 nm$

Hence, the wavelength of light used for the experiment is 400 nm.

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Given:
Separation between the two slits, $d = 1 mm = 10^{- 3} m$
Wavelength of the light used, $λ = 5.0 \times 10^{- 7} m$
Distance between screen and slit, $D = 1 m$
(a) The distance of the centre of the first minimum from the centre of the central maximum, x = $\frac{width of central maxima}{2}$
That is, $x = \frac{β}{2} = \frac{λ D}{2 d}$ ...(i)
$= \frac{5 \times 10^{- 7} \times 1}{2 \times 10^{- 3}} = 2.5 \times 10^{- 4} m = 0.25 mm$
(b) From equation (i),
fringe width, $β = 2 \times x = 0.50 mm$
So, number of bright fringes formed in one centimetre (10 mm) = $\frac{10}{0.50} = 20$ .

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Answer:

Given:
Separation between two narrow slits, $d = 0.8 mm = 0.8 \times 10^{- 3} m$
Wavelength of the yellow light, $λ = 589 nm = 589 \times 10^{- 9} m$
Distance between screen and slit, $D = 2.0 m$
Separation between the adjacent bright bands = width of one dark fringe
That is, $β = \frac{λ D}{d}$ ...(i)
$\Rightarrow β = \frac{589 \times 10^{- 9} \times 2}{0.8 \times 10^{- 3}} = 1.47 \times 10^{- 3} m = 1.47 mm .$

Hence, the adjacent bright bands in the interference pattern are 1.47 mm apart.

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Answer:

Given:
Wavelength of the blue-green light, $λ = 500 \times 10^{- 9} m$
Separation between two slits, $d = 2 \times 10^{- 3} m,$
Let angular separation between the consecutive bright fringes be θ.
$Using θ = \frac{β}{D} = \frac{λ D}{d D} = \frac{λ}{d}, we get: θ = \frac{500 \times 10^{- 9}}{2 \times 10^{- 3}} = 250 \times 10^{- 6} = 25 \times 10^{- 5} radian or 0.014 °$
Hence, the angular separation between the consecutive bright fringes is 0.014 degree.

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Answer:

Given:
Wavelengths of the source of light,
$λ_{1} = 480 \times 10^{- 9} m and λ_{2} = 600 \times 10^{- 9} m$
Separation between the slits, $d = 0.25 mm = 0.25 \times 10^{- 3} m$
Distance between screen and slit, $D = 150 cm = 1.5 m$
We know that the position of the first maximum is given by
$y = \frac{λ D}{d}$
So, the linear separation between the first maximum (next to the central maximum) corresponding to the two wavelengths = y₂ − y₁
$y_{2} - y_{1} = \frac{D (y_{2} - y_{1})}{d}$

$\Rightarrow y_{2} - y_{1} = \frac{1.5}{0.25 \times 10^{- 3}} (600 \times 10^{- 9} - 480 \times 10^{- 9}) y_{2} - y_{1} = 72 \times 10^{- 5} m = 0.72 mm$

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Answer:

Let the separation between the slits be d and distance between screen from the slits be D.
Suppose, the m^th bright fringe of violet light overlaps with the n^th bright fringe of red light.
Now, the position of the m^th bright fringe of violet light, y_v = $\frac{m λ_{v} D}{d}$
Position of the n^th bright fringe of red light, y_r = $\frac{n λ_{r} D}{d}$
For overlapping, y_{v =} y_r .
So, as per the question,
$\frac{m \times 400 \times D}{d} = \frac{n \times 700 \times D}{d} \Rightarrow \frac{m}{n} = \frac{7}{4}$
Therefore, the $7^{th}$ bright fringe of violet light overlaps with the 4^th bright fringe of red light.
It can also be seen that the 14^th violet fringe will overlap with the 8^th red fringe.
Because, $\frac{m}{n} = \frac{7}{4} = \frac{14}{8}$

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Answer:

Given:
The refractive index of the plate is $μ$ .
Let the thickness of the plate be 't' to produce a change in the optical path difference of $\frac{λ}{2}$ .
We know that optical path difference is given by $(μ - 1) t$ .
$∴ (μ - 1) t = \frac{λ}{2} \Rightarrow t = \frac{λ}{2 (μ - 1)}$
Hence, the thickness of a plate is $\frac{λ}{2 (μ - 1)}$ .

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Answer:

Given:
Refractive index of the plate is μ.
The thickness of the plate is t.
Wavelength of the light is λ.
(a)
When the plate is placed in front of the slit, then the optical path difference is given by $(μ - 1) t$ .

(b) For zero intensity at the centre of the fringe pattern, there should be distractive interference at the centre.
So, the optical path difference should be = $\frac{λ}{2}$
$i . e . (μ - 1) t = \frac{λ}{2} \Rightarrow t = \frac{λ}{2 (μ - 1)}$

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Answer:

Given:
Refractive index of the paper, μ = 1.45
The thickness of the plate, $t = 0.02 mm = 0.02 \times 10^{- 3} m$
Wavelength of the light, $λ = 620 nm = 620 \times 10^{- 9} m$
We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by $(μ - 1) t$ .
And optical path should be changed by λ for the shift of one fringe.
∴ Number of fringes crossing through the centre is
$n = \frac{(μ - 1) t}{λ} = \frac{(1.45 - 1) \times 0.02 \times 10^{- 3}}{620 \times 10^{- 9}} = 14.5$

Hence, 14.5 fringes will cross through the centre if the paper is removed.

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Answer:

Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate, $t = 1.964 micron = 1.964 \times 10^{- 6} m$
Let the wavelength of the light used = λ.
Number of fringes shifted is given by
$n = \frac{(μ - 1) t}{λ}$
So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
$Shift = n \times β = \frac{(μ - 1) t}{λ} \times \frac{λ D}{d} = \frac{(μ - 1) t \times D}{d} . . . (i)$
As per the question, when the distance between the screen and the slits is doubled,
i.e. $D' = 2 D$ ,
fringe width, $β = \frac{λ D'}{d} = \frac{λ 2 D}{d}$
According to the question, fringe shift in first case = fringe width in second case.
$So, \frac{(μ - 1) t \times D}{d} = \frac{λ 2 D}{d} \Rightarrow λ = \frac{(μ - 1) t}{2} = \frac{(1.6 - 1) \times (1.964) \times 10^{- 6}}{2} = 589.2 \times 10^{- 9} = 589.2 nm$

Hence, the required wavelength of the monochromatic light is 589.2 nm.

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Answer:

Given:
The thickness of the strips = $t_{1} = t_{2} = t = 0.5 mm = 0.5 \times 10^{- 3} m$
Separation between the two slits, $d = 0.12 cm = 12 \times 10^{- 4} m$
The refractive index of mica, μ_m = 1.58 and of polystyrene, μ_p = 1.58
Wavelength of the light, $λ = 590 nm = 590 \times 10^{- 9} m,$
Distance between screen and slit, D = 1 m

(a)
We know that fringe width is given by
$β = \frac{λ D}{d}$
$\Rightarrow β = \frac{590 \times 10^{- 9} \times 1}{12 \times 10^{- 4}} = 4.9 \times 10^{- 4} m$

(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
$∆ x = (μ_{m} - 1) t - (μ_{p} - 1) t = (μ_{m} - μ_{p}) t = (1.58 - 1.55) \times (0.5) (10^{- 3}) = (0.015) \times 10^{- 3} m$
∴ Number of fringes shifted, n = $\frac{∆ x}{λ}$ .
$\Rightarrow n = \frac{0.015 \times 10^{- 3}}{590 \times 10^{- 9}} = 25.43$
∴ 25 fringes and 0.43^th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43^th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
$x = (0.43) \times 4.91 \times 10^{- 4} (∵ β = 4.91 \times 10^{- 4} m) = 0.021 cm$
On the other side,
$x' = (1 - 0.43) \times 4.91 \times 10^{- 4} = 0.028 cm$

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Answer:

Given:
Refractive index of the two slabs are µ₁ and µ₂.
Thickness of both the plates is t.
When both the strips are fitted, the optical path changes by
$∆ x = (μ_{1} - 1) t - (μ_{2} - 1) t = (μ_{1} - μ_{2}) t$
For minimum at P₀, the path difference should be $\frac{λ}{2}$ .
$i . e . ∆ x = \frac{λ}{2}$
$So, \frac{λ}{2} = (μ_{1} - μ_{2}) t \Rightarrow t = \frac{λ}{2 (μ_{1} - μ_{2})}$
Therefore, minimum at point P₀ is $\frac{λ}{2 (μ_{1} - μ_{2})}$ .

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Answer:

Given:
The thickness of the thin paper, $t = 0.02 mm = 0.02 \times 10^{- 3} m$
Refractive index of the paper, $μ = 1.45$ .
Wavelength of the light, $λ = 600 nm = 600 \times 10^{- 9} m$
(a)
Let the intensity of the source without paper = I₁
and intensity of source with paper =I₂
Let a₁ and a₂ be corresponding amplitudes.
As per the question,
$I_{2} = \frac{4}{9} I_{1}$
We know that
$\frac{I_{1}}{I_{2}} = \frac{{a_{1}}^{2}}{{a_{2}}^{2}} (∵ I \propto a^{2}) \Rightarrow \frac{a_{1}}{a_{2}} = \frac{3}{2}$
Here, a is the amplitude.
$We know that \frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} . \Rightarrow \frac{I_{\max}}{I_{\min}} = \frac{{(3 + 2)}^{2}}{{(3 - 2)}^{2}} = \frac{25}{1} \Rightarrow I_{\max} : I_{\min} = 25 : 1$

(b)
Number of fringes that will cross through the centre is given by $n = \frac{(μ - 1) t}{λ}$ .
$\Rightarrow n = \frac{(1.45 - 1) \times 0.02 \times 10^{- 3}}{600 \times 10^{- 9}} = \frac{0.45 \times 0.02 \times 10^{4}}{6} = 15$

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Answer:

Given:
Separation between two slits, $d = 0.28 mm = 0.28 \times 10^{- 3} m$
Distance between screen and slit (D) = 48 cm = 0.48 m
Wavelength of the red light, $λ_{a} = 700 nm in vaccum = 700 \times 10^{- 9} m$
Let the wavelength of red light in water = $λ_{ω}$
We known that refractive index of water (μ_w =4/3),
μ_w = $\frac{Speed of light in vacuum}{Speed of light in the water}$
$So, μ_{w} = \frac{v_{a}}{v_{ω}} = \frac{λ_{a}}{λ_{ω}} \Rightarrow \frac{4}{3} = \frac{λ_{a}}{λ_{ω}} \Rightarrow λ_{ω} = \frac{3 λ_{a}}{4} = \frac{3 \times 700}{4} = 525 nm$
So, the fringe width of the pattern is given by
$β = \frac{λ_{ω} D}{d} = \frac{525 \times 10^{- 9} \times (0.48)}{(0.28) \times 10^{- 3}} = 9 \times 10^{- 4} = 0.90 mm$

Hence, fringe-width of the pattern formed on the screen is 0.90 mm.

Page No 381:

Answer:

Let the two slits are S₁ and S₂ with separation d as shown in figure.

The wave fronts reaching P₀from S₁ and S₂ will have a path difference of S₁X = ∆x.
In ∆S₁S₂X, $\sin θ = \frac{S_{1} X}{S_{1} S_{2}} = \frac{∆ x}{d}$
$\Rightarrow ∆ x = d \sin θ$
Using $θ = \sin^{- 1} (\frac{λ}{2 d})$ , we get,
$\Rightarrow ∆ x = d \times \frac{λ}{2 d} = \frac{λ}{2}$
Hence, there will be dark fringe at point P₀ as the path difference is an odd multiple of $\frac{λ}{2}$ .

Page No 381:

Answer:

(a) The phase of a light wave reflecting from a surface differs by ' $π$ ' from the light directly coming from the source.

Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of $π$ , which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero.

(b) Here, separation between two slits is $2 d$ .
Wavelength of the light is $λ$ .
Distance of the screen from the slit is $D$ .
Consider that the bright fringe is formed at position y. Then,
path difference, $∆ x = \frac{y \times 2 d}{D} = n λ$ .
After reflection from the mirror, path difference between two waves is $\frac{λ}{2}$ .
$\Rightarrow \frac{y \times 2 d}{D} = \frac{λ}{2} + n λ For first o r der, put n = 0 . \Rightarrow y = \frac{λ D}{4 d}$

Page No 382:

Answer:

Given:
Separation between two slits, $d' = 2 d = 2 mm = 2 \times 10^{- 3} m$ (as d = 1 mm)
Wavelength of the light used, $λ = 700 nm = 700 \times 10^{- 3} m$
Distance between the screen and slit (D) = 1.0 m
It is a case of Lloyd's mirror experiment.
$Fringe width, β = \frac{λD}{d'} = \frac{700 \times 10^{- 9} \times 1}{2 \times 10^{- 3}} = 0.35 \times 10^{- 3} m = 0.35 mm$

Hence, the width of the fringe is 0.35 mm.

Page No 382:

Answer:

Given:
The mirror reflects 64% of the energy or intensity of light.
Let intensity of source = I₁.
And intensity of light after reflection from the mirror = I₂.
Let a₁ and a₂ be corresponding amplitudes of intensity I₁ and I₂.
According to the question,
$I_{2} = \frac{I_{1} \times 64}{100} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{64}{100} = \frac{16}{25} And \frac{I_{2}}{I_{1}} = \frac{{a_{2}}^{2}}{{a_{1}}^{2}} \Rightarrow \frac{a_{2}}{a_{1}} = \frac{4}{5}$
$We know that \frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} = \frac{{(5 + 4)}^{2}}{{(5 - 4)}^{2}} I_{\max} : I_{\min} = 81 : 1$

Hence, the required ratio is 81 : 1.

Page No 382:

Answer:

Given:
Separation between the two slits = d
Wavelength of the coherent light =λ
Distance between the slit and mirror is D₁.
Distance between the slit and screen is D₂.
Therefore,
apparent distance of the screen from the slits,
$D = 2 D_{1} + D_{2} Fringe width, β = \frac{λ D}{d} = \frac{(2 D_{1} + D_{2}) λ}{d}$

Hence, the required fringe width is $\frac{(2 D_{1} + D_{2}) λ}{d}$ .

Page No 382:

Answer:

Given: Separation between two slits,
$d = 0.5 mm = 0.5 \times 10^{- 3} m$
Wavelength of the light, $λ = 400 nm to 700 nm$
Distance of the screen from the slit, $D = 50 cm = 0.5 m$ ,
Position of hole on the screen, $y_{n} = 1 mm = 1 \times 10^{- 3} m$

(a) The wavelength(s) will be absent in the light coming from the hole, which will form a dark fringe at the position of hole.
$y_{n} = \frac{(2 n + 1) λ_{n}}{2} \frac{D}{d}$ , where n = 0, 1, 2, ...

$\Rightarrow λ_{n} = \frac{2}{(2 n + 1)} \frac{y_{n} d}{D} = \frac{2}{(2 n + 1)} \times \frac{10^{- 3} \times 0.05 \times 10^{- 3}}{0.5} = \frac{2}{(2 n + 1)} \times 10^{- 6} m = \frac{2}{(2 n + 1)} \times 10^{3} nm$

$For n = 1, λ_{1} = (\frac{2}{3}) \times 1000 = 667 nm For n = 2, λ_{2} = (\frac{2}{5}) \times 1000 = 400 nm$
Thus, the light waves of wavelength 400 nm and 667 nm will be absent from the light coming from the hole.

(b) The wavelength(s) will have a strong intensity, which will form a bright fringe at the position of the hole.
$So, y_{n} = n λ_{n} \frac{D}{d} \Rightarrow λ_{n} = y_{n} \frac{d}{n D} For n = 1, λ_{1} = y_{n} \frac{d}{D} = 10^{- 3} \times (0.5) \times \frac{10^{- 3}}{0.5} = 10^{- 6} m = 1000 nm .$

But 1000 nm does not fall in the range 400 nm − 700 nm.

$Again, for n = 2, λ_{2} = y_{n} \frac{d}{2 D} = 500 nm$
So, the light of wavelength 500 nm will have strong intensity.

Page No 382:

Answer:

From the figure, $AB = BO and AC = CO$ .
Path difference of the wave front reaching O,
$∆ x = (AB + BO) - (AC + CO) = 2 (AB - AC) = 2 (\sqrt{D^{2} + d^{2}} - D)$
For dark fringe to be formed at O, path difference should be an odd multiple of $\frac{λ}{2}$ .
$So, ∆ x = (2 n + 1) \frac{λ}{2} \Rightarrow 2 (\sqrt{D^{2} + d^{2}} - D) = (2 n + 1) \frac{λ}{2} \Rightarrow \sqrt{D^{2} + d^{2}} = D + (2 n + 1) \frac{λ}{4} \Rightarrow D^{2} + d^{2} = D^{2} + {(2 n + 1)}^{2} \frac{λ^{2}}{16} + (2 n + 1) \frac{λ D}{2}$
Neglecting, ${(2 n + 1)}^{2} \frac{λ^{2}}{16}$ , as it is very small, we get:
$d = (\sqrt{2 n} + 1) \frac{λ D}{2} For minimum d, putting, n = 0 d_{\min} = \sqrt{(\frac{λ D}{2})}$ , we get:
Thus, for $d_{\min} = \sqrt{(\frac{λ D}{2})}$ there is a dark fringe at O.

Page No 382:

Answer:

Let P be the point of minimum intensity.
For minimum intensity at point P,

$S_{1} P - S_{2} P = x = (2 n + 1) \frac{λ}{2}$

Thus, we get:

$\sqrt{Z^{2} + {(2 λ)}^{2}} - Z = (2 n + 1) \frac{λ}{2} \Rightarrow Z^{2} + 4 λ^{2} = Z^{2} {(2 n + 1)}^{2} \frac{λ^{2}}{4} + 2 Z (2 n + 1) \frac{λ}{2} \Rightarrow Z = \frac{4 λ^{2} - (2 n + 1) 2 λ^{2} / 4}{(2 n + 1) λ} = \frac{16 λ^{2} - (2 n + 1)}{4 (2 n + 1) λ} . . . (i)$

$When n = 0, Z = \frac{15 λ}{4} n = - 1, Z = \frac{- 15 λ}{4} n = 1, Z = \frac{7 λ}{12} n = 2, Z = \frac{- 9 λ}{20}$

Thus, $Z = \frac{7 λ}{12}$ is the smallest distance for which there will be minimum intensity.

Page No 382:

Answer:

(a) Given:
Wavelength of light = $λ$
Path difference of wave fronts reaching from A and B is given by
$∆ x_{B} = {BP}_{0} - {AP}_{0} = \frac{λ}{3} \Rightarrow \sqrt{D^{2} + d^{2}} - D = \frac{λ}{3} \Rightarrow D^{2} + d^{2} = D^{2} + \frac{λ^{2}}{9} + \frac{2 λ D}{3}$

We will neglect the term $\frac{λ^{2}}{9}$ , as it has a very small value.
$∴ d = \sqrt{\frac{(2 λ D)}{3}}$

(b) To calculating the intensity at P₀_,consider the interference of light waves coming from all the three slits.
Path difference of the wave fronts reaching from A and C is given by
${CP}_{0} - {AP}_{0} = \sqrt{D^{2} + {(2 d)}^{2}} - D = \sqrt{D^{2} + \frac{8 λ D}{3}} - D (Using the value of d from part a) = D {\{1 + \frac{8 λ}{3 D}\}}^{\frac{1}{2}} - D Expanding the value using binomial theorem and neglecting the higher order terms, we get : D \{1 + \frac{1}{2} \times \frac{8 λ}{3 D} + . . .\} - D$
${CP}_{0} - {AP}_{0} = \frac{4 λ}{3}$
So, the corresponding phase difference between the wave fronts from A and C is given by
$ϕ_{c} = \frac{2 π ∆ x_{C}}{λ} = \frac{2 π \times 4 λ}{3 λ} \Rightarrow ϕ_{c} = \frac{8 π}{3} or (2 π + \frac{2 π}{3}) \Rightarrow ϕ_{c} = \frac{2 π}{3} . . . (1) Again, ϕ_{B} = \frac{2 π ∆ x_{B}}{λ} \Rightarrow ϕ_{B} = \frac{2 π λ}{3 λ} = \frac{2 π}{3} . . . (2)$

So, it can be said that light from B and C are in the same phase, as they have the same phase difference with respect to A.

Amplitude of wave reaching P₀ is given by
$A = \sqrt{{(2 a)}^{2} + a^{2} + 2 a \times a \cos (\frac{2 π}{3})} = \sqrt{4 a^{2} + a^{2} + 2 a^{2} \sqrt{3}} ∴ l_{p o} = K {(\sqrt{3} r)}^{2} = 3 K r^{2} = 3 l$
Here, I is the intensity due to the individual slits and I_po is the total intensity at P₀.
Thus, the resulting amplitude is three times the intensity due to the individual slits.

Page No 382:

Answer:

Given:
Separation between the slits, $d = 2 mm = 2 \times 10^{- 3} m$
Wavelength of the light, $λ = 600 nm = 6 \times 10^{- 7} m$
Distance of the screen from the slits, D = 2⋅0 m
$I_{\max} = 0.20 W / m^{2} For the point at a position y = 0.5 cm = 0.5 \times 10^{- 2} m, p ath difference, ∆ x = \frac{y d}{D} . \Rightarrow ∆ x = \frac{0.5 \times 10^{- 2} \times 2 \times 10^{- 3}}{2} = 5 \times 10^{- 6} m$

So, the corresponding phase difference is given by
$∆ ϕ = \frac{2 π ∆ x}{λ} = \frac{2 π \times 5 \times 10^{- 6}}{6 \times 10^{- 7}} = \frac{50 π}{3} = 16 π + \frac{2 π}{3} or ∆ ϕ = \frac{2 π}{3}$
So, the amplitude of the resulting wave at point y = 0.5 cm is given by
$A = \sqrt{a^{2} + a^{2} + 2 a^{2} \cos (\frac{2 π}{3})} = \sqrt{a^{2} + a^{2} - a^{2}} = a$
Similarly, the amplitude of the resulting wave at the centre is 2a.
Let the intensity of the resulting wave at point y = 0.5 cm be I.
$Since \frac{I}{I_{\max}} = \frac{A^{2}}{{(2 a)}^{2}}, we have : \frac{I}{0.2} = \frac{A^{2}}{4 a^{2}} = \frac{a^{2}}{4 a^{2}} \Rightarrow I = \frac{0.2}{4} = 0.05 W / m^{2}$
Thus, the intensity at a point 0.5 cm away from the centre along the width of the fringes is 0.05 W/m².

Page No 382:

Answer:

Given:
Separation between the two slits = d
Wavelength of the light = $λ$
Distance of the screen = $D$
(a) When the intensity is half the maximum:
Let I_max be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, $I = a^{2} + a^{2} + 2 a^{2} \cos ϕ$
Here, $ϕ$ is the phase difference in the waves coming from the two slits.
So, $I = 4 a^{2} \cos^{2} (\frac{ϕ}{2})$
$\Rightarrow \frac{I}{I_{\max}} = \frac{1}{2} \Rightarrow \frac{4 a^{2} \cos^{2} (\frac{ϕ}{2})}{4 a^{2}} = \frac{1}{2} \Rightarrow \cos^{2} (\frac{ϕ}{2}) = \frac{1}{2} \Rightarrow \cos (\frac{ϕ}{2}) = \frac{1}{\sqrt{2}} \Rightarrow \frac{ϕ}{2} = \frac{π}{4} \Rightarrow ϕ = \frac{π}{2} Corrosponding path difference, ∆ x = \frac{λ}{4} \Rightarrow y = \frac{∆ x D}{d} = \frac{λ D}{4 d}$

(b) When the intensity is one-fourth of the maximum:

$\frac{I}{I_{\max}} = \frac{1}{4} \Rightarrow 4 a^{2} \cos^{2} (\frac{ϕ}{2}) = \frac{1}{4} \Rightarrow \cos^{2} (\frac{ϕ}{2}) = \frac{1}{4} \Rightarrow \cos (\frac{ϕ}{2}) = \frac{1}{2} \Rightarrow \frac{ϕ}{2} = \frac{π}{3} So, corrosponding path difference, ∆ x = \frac{λ}{3} and position, y = \frac{∆ x D}{d} = \frac{λ D}{3 d} .$

Page No 382:

Answer:

Given:
Separation between the two slits, $d = 1 mm = 10^{- 3} m$
Wavelength of the light, $λ = 500 nm = 5 \times 10^{- 7} m$
Distance of the screen, $D = 1 m$
Let I_max be the maximum intensity and I be the intensity at the required point at a distance y from the central point.
So, $I = a^{2} + a^{2} + 2 a^{2} \cos ϕ$
Here, $ϕ$ is the phase difference in the waves coming from the two slits.
So, $I = 4 a^{2} \cos^{2} (\frac{ϕ}{2})$
$\Rightarrow \frac{I}{I_{\max}} = \frac{1}{2} \Rightarrow \frac{4 a^{2} \cos^{2} (\frac{ϕ}{2})}{4 a^{2}} = \frac{1}{2} \Rightarrow \cos^{2} (\frac{ϕ}{2}) = \frac{1}{2} \Rightarrow \cos (\frac{ϕ}{2}) = \frac{1}{\sqrt{2}} \Rightarrow \frac{ϕ}{2} = \frac{π}{4} \Rightarrow ϕ = \frac{π}{2} Corrosponding path difference, ∆ x = \frac{1}{4} \Rightarrow y = \frac{∆ x D}{d} = \frac{λ D}{4 d}$
$\Rightarrow y = \frac{5 \times 10^{- 7} \times 1}{4 \times 10^{- 3}} = 1.25 \times 10^{- 4} m$

∴ The required minimum distance from the central maximum is $1.25 \times 10^{- 4} m$ .

Page No 382:

Answer:

Given:
Separation between two slits = d
Wavelength of the light = $λ$
Distance of the screen = $D$
Let I_max be the maximum intensity and I be half the maximum intensity at a point at a distance y from the central point.
So, $I = a^{2} + a^{2} + 2 a^{2} \cos ϕ$
Here, $ϕ$ is the phase difference in the waves coming from the two slits.
So, $I = 4 a^{2} \cos^{2} (\frac{ϕ}{2})$
$\Rightarrow \frac{I}{I_{\max}} = \frac{1}{2} \Rightarrow \frac{4 a^{2} \cos^{2} (\frac{ϕ}{2})}{4 a^{2}} = \frac{1}{2} \Rightarrow \cos^{2} (\frac{ϕ}{2}) = \frac{1}{2} \Rightarrow \cos (\frac{ϕ}{2}) = \frac{1}{\sqrt{2}} \Rightarrow \frac{ϕ}{2} = \frac{π}{4} \Rightarrow ϕ = \frac{π}{2} Corrosponding path difference, ∆ x = \frac{1}{4} \Rightarrow y = \frac{∆ x D}{d} = \frac{λ D}{4 d}$
The line-width of a bright fringe is defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
So, line-width = 2y
$= 2 \frac{D λ}{4 d} = \frac{D λ}{2 d}$

Thus, the required line width of the bright fringe is $\frac{D λ}{2 d}$ .

Page No 383:

Answer:

Given:
Separation between the two slits = d
Wavelength of the light = $λ$
Distance of the screen = $D$
The fringe width (β) is given by $β = \frac{λ D}{d}$ .
At S₃, the path difference is zero. So, the maximum intensity occurs at amplitude = 2a.
(a) When $z = \frac{D λ}{2 d}$ :
The first minima occurs at S₄, as shown in figure (a).
With amplitude = 0 on screen ∑₂_,we get:
$\frac{l_{m a x}}{l_{m i n}} = \frac{{(2 a + 0)}^{2}}{{(2 a - 0)}^{2}} = 1$

(b) When $z = \frac{D λ}{d}$ :
The first maxima occurs at S₄, as shown in the figure.

With amplitude = $2 a$ on screen ∑₂, we get:
$\frac{l_{\max}}{l_{\min}} = \frac{{(2 a + 2 a)}^{2}}{{(2 a - 2 a)}^{2}} = \infty$

(c) When $z = \frac{D λ}{4 d}$ :

The slit S₄ falls at the mid-point of the central maxima and the first minima, as shown in the figure.
$Intensity = \frac{l_{\max}}{2} \Rightarrow Amplitude = \sqrt{2} a$
$∴ \frac{l_{\max}}{l_{\min}} = \frac{{(2 a + \sqrt{2} a)}^{2}}{{(2 a - \sqrt{2} a)}^{2}} = 34$

Page No 383:

Answer:

Given:
Fours slits S₁, S₂, S₃ and S₄.
The separation between slits S₃ and S₄ can be changed.
Point P is the common perpendicular bisector of S₁S₂ and S₃S₄.

(a) $For z = \frac{λ D}{d}$ :
The position of the slits from the central point of the first screen is given by
$y = {OS}_{3} = {OS}_{4} = \frac{z}{2} = \frac{λ D}{2 d}$
The corresponding path difference in wave fronts reaching S₃ is given by
$∆ x = \frac{y d}{D} = \frac{λ D}{2 d} \times \frac{d}{D} = \frac{λ}{2}$
Similarly at S₄, path difference, $∆ x = \frac{y d}{D} = \frac{λ D}{2 d} \times \frac{d}{D} = \frac{λ}{2}$
$i . e . dark fringes are formed at S_{3} and S_{4}$ .
So, the intensity of light at S₃ and S₄ is zero. Hence, the intensity at P is also zero.

(b) $For z = \frac{3 λ D}{2 d}$
The position of the slits from the central point of the first screen is given by
$y = {OS}_{3} = {OS}_{4} = \frac{z}{2} = \frac{3 λ D}{4 d}$
The corresponding path difference in wave fronts reaching S₃ is given by
$∆ x = \frac{y d}{D} = \frac{3 λ D}{4 d} \times \frac{d}{D} = \frac{3 λ}{4}$
Similarly at S₄, path difference, $∆ x = \frac{y d}{D} = \frac{3 λ D}{4 d} \times \frac{d}{D} = \frac{3 λ}{4}$
Hence, the intensity at P is I.

(c) Similarly, for $z = \frac{2 D λ}{d}$ ,
the intensity at P is 2I.

Page No 383:

Answer:

Given:
Thickness of soap film, d =0.0011 mm = 0.0011 × 10⁻³ m
Wavelength of light used, $λ = 580 nm = 580 \times 10^{- 9} m$
Let the index of refraction of the soap solution be μ.
The condition of minimum reflection of light is 2μd = nλ,
where n is an interger = 1 , 2 , 3 ...
$\Rightarrow μ = \frac{n λ}{2 d} = \frac{2 n λ}{4 d} = \frac{580 \times 10^{- 9} \times (2 n)}{4 \times 11 \times 10^{- 7}} = \frac{5.8 (2 n)}{44} = 0.132 (2 n)$
As per the question, μ has a value between 1.2 and 1.5. So,
$n = 5 So, μ = 0.132 \times 10 = 1.32$

Therefore, the index of refraction of the soap solution is 1.32.

Page No 383:

Answer:

Given:
Wavelength of light used, $λ = 560 \times 10^{- 9} m$
Refractive index of the oil film, $μ = 1.4$
Let the thickness of the film for strong reflection be t.
The condition for strong reflection is
$2 μ t = (2 n + 1) \frac{λ}{2} \Rightarrow t = (2 n + 1) \frac{λ}{4 μ}$
where n is an integer.
For minimum thickness, putting n = 0, we get:
$t = \frac{λ}{4 μ} = \frac{560 \times 10^{- 9}}{4 \times 1.4} = 10^{- 7} m = 100 nm$
Therefore, the minimum thickness of the oil film so that it strongly reflects the light is 100 nm.

Page No 383:

Answer:

Given,
Wavelength of light used, $λ = 400 \times 10^{- 9} m to 700 \times 10^{- 9} nm$
Refractive index of water, $μ = 1.33$
The thickness of film, $t = 10^{- 4} cm = 10^{- 6} m$
The condition for strong transmission: $2 μ t = n λ$ ,
where n is an integer.
$\Rightarrow λ = \frac{2 μ t}{n}$
$\Rightarrow λ = \frac{2 \times 1.33 \times 10^{- 6}}{n} = \frac{2660 \times 10^{- 9}}{n} m$
Putting n = 4, we get, λ₁ = 665 nm.
Putting n = 5, we get, λ₂ = 532 nm.
Putting n = 6, we get, λ₃ = 443 nm.
Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.

Page No 383:

Answer:

Given:
Wavelength of light used, $λ = 400 \times 10^{- 9} to 750 \times 10^{- 9} m$
Refractive index of oil, μ_oil, is 1.25 and that of glass, μ_g, is 1.50.
The thickness of the oil film, $d = 1 \times 10^{- 4} cm = 10^{- 6} m,$
The condition for the wavelengths which can be completely transmitted through the oil film is given by
$λ = \frac{2 μ d}{(n + \frac{1}{2})} = \frac{2 \times 10^{- 6} \times (1.25) \times 2}{(2 n + 1)} = \frac{5 \times 10^{- 6}}{(2 n + 1)} m \Rightarrow λ = \frac{5000}{(2 n + 1)} nm$
Where n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,
$λ = \frac{5000}{(2 \times 3 + 1)} = \frac{5000}{7} = 714.3 nm$
When, n = 4, we get,
$λ = \frac{5000}{(2 \times 4 + 1)} = \frac{5000}{9} = 555.6 nm$
When, n = 5, we get,
$λ = \frac{5000}{(2 \times 5 + 1)} = \frac{5000}{11} = 454.5 nm$
Thus the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.

Page No 383:

Answer:

Given:
Width of the slit, b = 5.0 cm
First diffraction minimum is formed at θ = 30°.
For the diffraction minima, we have:
bsinθ = nλ
For the first minima, we put n = 1.
$5 \times \sin 30 ° = 1 \times λ \Rightarrow λ = \frac{5}{2} = 2.5 cm$
Therefore, the wavelength of the microwaves is 2.5 cm.

Page No 383:

Answer:

Given:
Wavelength of the light used, $λ = 560 nm = 560 \times 10^{- 9} m$
Diameter of the pinhole, d = 0.20 mm = 2 × 10⁻⁴ m
Distance of the wall, D = 2m
We know that the radius of the central bright spot is given by
$R = 1.22 \frac{λ D}{d} = 1.22 \times \frac{560 \times 10^{- 9} \times 2}{2 \times 10^{- 4}} = 6.832 \times 10^{- 3} m or = 0.683 cm$

Hence, the diameter 2R of the central bright spot on the wall is 1.37 cm.

Page No 383:

Answer:

Given:
Wavelength of the light used, $λ = 620 nm = 620 \times 10^{- 9} m$
Diameter of the convex lens, $d = 8 cm = 8 \times 10^{- 2} m$
Distance from the lens where light is to be focused, $D = 20 cm = 20 \times 10^{- 2} m$
The radius of the central bright spot is given by
$R = 1.22 \frac{λ D}{d} = 1.22 \times \frac{620 \times 10^{- 9} \times 20 \times 10^{- 2}}{8 \times 10^{- 2}} = 1891 \times 10^{- 9} m \approx 1.9 \times 10^{- 6} m$
∴ Diameter of the central bright spot, 2R = 3.8 × 10⁻⁶ m.

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