Hc Verma I for Class 12 Science Physics Chapter 22 - Photometry

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Page No 453:

Answer:

The luminous flux of a source emitting radio waves will be zero, as the luminosity of radio waves is zero.

Page No 453:

Answer:

The luminous flux of a 1 W sodium vapour lamp is more than that of a 10 kW source of ultraviolet radiation. This is because the luminosity of a 1 W sodium vapour lamp is 589 nm or 589.6 nm, whereas the luminosity of UV radiation is zero.

Page No 453:

Answer:

If the surface is rotated by 30⁰, the illuminance will decrease. This is because illuminance depends upon the cosine of the normal angle.
Yes, if the light does not fall normally on the surface initially, it may increase or decrease depending upon the former's angle. If the 30⁰ rotation brings the table closer to the normal of the surface, the illuminance will increase; otherwise, it will decrease.

Page No 453:

Answer:

The illuminance will be maximum on the area just below the bulb that is normal to the table.
Yes, the plane mirror will increase the radiant flux on the table, which will further increase the luminance flux of the table. Therefore, illuminance will increase.

Page No 453:

Answer:

During noon, the Sun's rays fall directly on the Earth's surface. The Sun being a yellow star, its rays also appear yellow at noon. Since yellow light has a high relative luminosity, it produces a high sensation of visibility in our eyes, thereby making the Sun appear brighter.
However, in the morning and evening, due to the slanting rays of the Sun on the Earth's surface, the smaller and middle range wavelength of the rays get scattered in the upper atmosphere. Thus, the sun appears orange during these two times. Since red/orange light has a low relative luminosity, it produces a low sensation of visibility in our eyes, thereby making the Sun appear less brighter.

Page No 453:

Answer:

In a filament bulb, heat energy is converted into light energy. Most of the electrical energy supplied to the bulb is radiated as heat and only a small percentage is radiated as visible light. Since heat waves are not visible, they don't contribute to visibility. Thus, its luminous flux is lesser than the total radiant flux; hence, its luminous efficiency is low.

In a mercury vapour lamp, a greater amount of electrical energy supplied is converted into visible radiation. Thus, its luminous flux is relatively higher than incandescent lamps.

Page No 453:

Answer:

No, we cannot increase the illuminating power of any white light source by wrapping a yellow plastic around them.
White light emits power in all wavelengths of visible radiation, and the yellow plastic will block all the other wavelengths other than the yellow one. This means that it will allow only a small amount of power, emitted by the white light, in the yellow coloured wavelength and block the rest. Thus, it will decrease the illuminating power of the source.

Page No 454:

Answer:

Correct option (d)

Total luminous flux is the total brightness producing capability of a radiating source. Or, it is the measurement of the total energy entering our eyes that produces the sensation of vision.

(a) This cannot be the correct answer because all energies cannot be sensed by our eyes.
(b) This cannot be the correct answer because all wavelengths do not produce any sensation in our eyes.
(c) This cannot be the correct answer because all wavelengths contributing the radiant flux are not always visible to our eyes. There may be a large radiant flux yet there may not be any sensation of vision.

Page No 454:

Answer:

Correct option (c)

Wavelength of light B is 555 nm. It has the highest luminosity; hence, X_B will be highest.
Again, 450 nm is nearer to 555 nm than 700 nm.
∴ 555 - 450 = 105
But 700 - 555 = 145
So, X_A's brightness will be greater than that of X_C_.

Page No 454:

Answer:

The correct option is (c).

The luminosity first increases up to 555 nm and then decreases.

Page No 454:

Answer:

Correct option (b)

$Here, r = \sqrt{2} tanθ = \frac{BC}{AB} = 1 I_{o} = 40 lux θ = \tan^{- 1} (1) = 45^{0} The illuminance is given by E = \frac{I_{o} cosθ}{r^{2}} = \frac{40 \times \cos (45^{0})}{{(\sqrt{2})}^{2}} = 14 lux$

Page No 454:

Answer:

Correct option (c)
Illuminance is given by:
$E = \frac{I_{o} \cos θ}{r^{2}}$
$θ = 0^{0} \frac{Δ r}{r} = 1 % E = \frac{I_{o}}{r^{2}} Differentiating, d E = - 2 \frac{I_{o}}{r^{3}} d r As approximation differentials are replaced by Δ, ΔE = - 2 \frac{I_{o}}{r^{3}} Δr \Rightarrow ΔE = - 2 \frac{I_{o}}{r^{2}} (\frac{Δr}{r}) \Rightarrow ΔE = - 2 E (\frac{Δr}{r}) \Rightarrow \frac{ΔE}{E} = - 2 (\frac{Δr}{r}) \Rightarrow \frac{ΔE}{E} = - 2 \times 1 % = - 2 % S i n c e, n egative sign implies decrease; h e n c e, i l l u m i n a n c e d e c r e a s e s b y 2 % .$

Page No 454:

Answer:

Correct option (a)

Since the beam is parallel, it will have no angular spread. So the illuminance will remain same throughout. Therefore, in this case, it will be 40 lux.

Page No 454:

Answer:

Correct option (c).

Let us consider two coaxial cylindrical surfaces at distances r and r' from the axis. Let areas dA and dA' subtend the solid angle dω at the central axis. The height of the area element will be same, i.e. equal to dy. Let the breath of dA be dx and that of dA' be dx'.
$Now from the arc s, dx = rdθ dx' = r' dθ Now, dA = dxdy = rdθdy dA' = dx' dy = r' dθdy \frac{dA}{dA'} = \frac{r}{r'} \Rightarrow \frac{dA}{r} = \frac{dA'}{r'} = dω$

The luminous flux going through the solid angle dω will be:
dF = Idω

$Now, dF = I \frac{dA}{r} If the surfaces are inclined at an angle α, dF = I \frac{dAcosα}{r} Now, illuminance is defined as E = \frac{dF}{dA} = I \frac{dAcosα}{r} \Rightarrow E \propto \frac{1}{r}$

Page No 454:

Answer:

Correct option (b)

Here,

$d_{1} = 5 c m = 0.05 m d_{2} = 10 c m = 0.1 m t_{1} = 3 s t_{2} = ? Let t h e actual incident illuminance b e E_{o} Let t h e iluminance a t 3 cm distance b e E_{d 1} Let t h e iluminance a t 10 cm distance b e E_{d 2} \cos θ = 1 E_{d 1} = \frac{E_{o}}{{d_{1}}^{2}} Now, t_{1} α \frac{1}{E_{d 1}} \Rightarrow t_{1} = \frac{k}{E_{d 1}} \Rightarrow t_{1} = \frac{k 5^{2}}{E_{o}} \Rightarrow \frac{k}{E_{o}} = \frac{3}{25} S imilarly, \Rightarrow t_{2} = \frac{k 10^{2}}{E_{o}} \Rightarrow t_{2} = \frac{3}{25} \times 10^{2} = 12 s$

Page No 454:

Answer:

Correct option (c)

$Here, t_{1} = 12 s θ_{1} = 0^{0} θ_{2} = 60^{0} t_{2} = ? L et t h e distance b e r . L et t h e incident luminosity b e E_{o} . We have, E_{θ 1} = \frac{E_{o} {cosθ}_{1}}{r^{2}} t_{1} α \frac{1}{E_{θ 1}} \Rightarrow t_{1} = \frac{r^{2} k}{E_{o} {cosθ}_{1}} \Rightarrow 12 = \frac{r^{2} k}{E_{o} \cos 0} \Rightarrow \frac{r^{2} k}{E_{o}} = 12 S imilarly, t_{2} = \frac{r^{2} k}{E_{o} {cosθ}_{2}} = \frac{12}{\cos (60^{0})} \Rightarrow t_{2} = 12 \times 2 = 24 s$

Page No 454:

Answer:

Correct option (c)

Let the distance between the parallel straight lines be L.
Angle with normal = θ
$We know, I = \frac{I_{o} \cos θ}{r^{2}} From t h e a b o v e fig u r e, w e g e t I = \frac{I_{o} \cos (90^{0} - α)}{r^{2}} \Rightarrow I = \frac{I_{o} sinα}{r^{2}} \Rightarrow I = \frac{I_{o}}{r^{2}} (\frac{L}{r}) \Rightarrow I = \frac{I_{o} L}{r^{3}} L = constant for parallel moving source S o, I_{o} L = k (constant) \Rightarrow I = \frac{k}{r^{3}} \Rightarrow Iα \frac{1}{r^{3}}$

Page No 454:

Answer:

Correct option (d)

Here, since B and C have the same distance and lie inclined to make the the same vertical angle with the source, they will have same intensity. But, as A is nearer and forms an angle of 0⁰ with the source, intensity of A will be greater than C or B.

Page No 454:

Answer:

Correct option (c) and (d)

Brightness depends upon how our eyes perceive light. Our eyes perceive yellow colour the most, so brightness depends upon the colour of the source. Now, colour is related to the wavelength of the source; so, brightness depends upon the wavelength as well.
Our eyes detect brightness by the amount of photons actually reaching our retinas. Again, the number of photons depends upon the power of the source. So, brightness depends upon the power of the source too.

Page No 454:

Answer:

Correct option (a), (b), (c), (d) i.e. all the options are correct.
a) Moving the source to the middle will illuminate the surface properly because illuminance depends upon the distance from the source.
b) Rotating the source will also have an effect because illuminance depends upon the angle made by the normal on the surface.
c) Bringing mirrors to the proper position will increase illuminance at that particular portion of the wall by gathering light and focussing them at one point.
d) Our eyes sense some colours as bright and some colours as dull, selecting the colours near yellow will make the wall appear brighter.

Page No 454:

Answer:

Correct options: (b) and (c)

a) The luminous efficiency of a monochromatic source may be less than that of the white light if the former emits wavelength far away from 555 nm.
b) Yes, it is true that our eyes mostly respond to colours close to the wavelength of 555 nm and detect them bright. So, luminous efficiency is unity (highest).
c) It is true because white light distributes its energy amongst certain colours that our eyes cannot detect as brightly as they detect a 555 nm light.
d) It is not necessarily true. If the monochromatic light radiates in a wavelength that is far away from 555 nm, our eyes will not perceive it as bright. So, it will have lesser illuminating power.

Page No 454:

Answer:

Correct option (b), (c), (d)

a) No, luminous flux has the dimension of luminous intensity (cd/sr). The dimension of radiant flux is watt.
b) Yes, both have the dimension of luminous intensity, i.e. cd/sr.
c) Yes, both have the dimensions of power.
d) Yes, it is a ratio of same kind of quantities. So, it is dimensionless.

Page No 455:

Answer:

Given,
Total energy emitted (E) = 45 J
Time (t) = 15 s
Radiant flux of the source is given by,
$Radiant Flux = \frac{Total energy emitted}{Time}$
$= \frac{45}{15} = 3 W$
So, the radiant flux of the source is 3 W.

Page No 455:

Answer:

Let t be the time for which the photograph is exposed.
To record the sufficiently intense lines, energy should be same.
Energy = radiant flux × time
= 10 W×12 s
= 12 W×t
$\Rightarrow t = \frac{10 W \times 12 s}{12 W} = 10 s$
Therefore, the photographic plate should be exposed for 10 s to get equally intense lines.

Page No 455:

Answer:

From the graph, we can find the following:
(a) The relative luminosity of wavelength 480 nm is 0.14.
(b) The relative luminosity of wavelength 520 nm is 0.68.
(c) The relative luminosity of wavelength 580 nm is 0.92.
(d) The relative luminosity of wavelength 600 nm is 0.66.

Page No 455:

Answer:

Given,
Relative luminosity = 0.6
Let P be the radiant flux of the source.
∴ Luminous flux = P $\times$ 685
Again, relative luminosity $= \frac{Luminous flux of a source of given wavelength}{Luminous flux of a source of 555 nm of same power}$

$∴ 0.6 = \frac{Luminous flux of a source P watt}{685 P}$
$\Rightarrow 685 \times P \times 0.6 = 120 \times 685$
$\Rightarrow P = \frac{120}{0.6} = 200 W$
So, 200 W radiant flux is needed to produce the same brightness sensation for 600 nm wavelength.

Page No 455:

Answer:

Given,
The luminous flux of the given source is 450 lumen/watt.
$We know that luminous flux of 555 nm source of 1 W = 685 lum e n s$
∴ Relative luminosity is given by,

$\frac{Luminous flux of the source of given wavelength}{Luminous flux of 555 nm source of same power} = \frac{450}{685} = 66 %$

So, the relative luminosity at the wavelength emitted is 66 %.

Page No 455:

Answer:

Given,
The radiant flux of the light of wavelength 555 nm is 40 W.
The radiant flux of the light of wavelength 600 nm is 30 W.
The relative luminosity at 600 nm is 0.6.

(a) Total radiant flux = radiant flux of 555 nm part of light + radiant flux of 600 nm part of light

                                 = 40 W + 30 W = 70 W

(b) Total luminous flux = luminous flux of 555 nm part of light + luminous flux of 600 nm part of light
                                 = 1 × 40 × 685 + 0.6 × 30 × 685
                                     = 39730 lumen

(c)
$Luminous efficiency = \frac{Total luminous flux}{Total radiant flux} = \frac{39730}{70} = 567.6 lumen / W$
So, the luminous efficiency is 568 lumen/W.

Page No 455:

Answer:

Given:
Wavelength of light, $λ$ = 555 nm
Radiant flux emitted = 35 W
Power input = 100 W
Overall luminous efficiency = $\frac{Total luminous flux}{Power input} = \frac{35 \times 685}{100} = 239.75 lm / W$
So, the overall luminous efficiency will be 240 lm/W.

Page No 455:

Answer:

Given that,
Radiant flux = 31.4 W
Since the radiant flux is distributed uniformly in all directions, the solid angle will be 4 $π$
Luminous efficiency = 60 lumen/W
So, luminous flux = luminous efficiency $\times$ radiant flux
                            = 60 × 31.4 lumen

Luminous intensity = $\frac{Luminous Flux}{Solid angle} = \frac{Luminous Flux}{4 π}$
      $= \frac{60 \times (31.4)}{4 π}$
                            $= 150 candela$

Page No 455:

Answer:

Given,
Luminous flux = 628 lumen
Angle made by the normal with the x axis ( $θ$ ) = 37^ₒ
Distance of point, r = 1 m
Since the radiant flux is distributed uniformly in all directions, the solid angle will be 4 $π$ .
∴ Luminous intensity, l = $\frac{Luminous Flux}{Solid angle}$
$= \frac{628}{4 π} = 50 candela$

$I lluminance (E) is given by, E = l \cos \frac{θ}{r^{2}}$
On substituting the respective values we get,
$E = 50 \times \frac{\cos 37 °}{1^{2}} = 50 \times \frac{(4 / 5)}{1} = 40 lux$
So, the illuminance on the area is 40 lux.

Page No 455:

Answer:

Let the luminous intensity of the source be l and the distance between the source and the area, in the initial position, be x.
Given,
Initial illuminance (E_A) = 900 lumen/m²
Final illuminance (E_B) = 400 lumen/m²
Illuminance on the initial position is given by,
$E_{A} = l \frac{\cos θ}{x^{2}}$ ......(1)
Illuminance at final position is given by
$E_{B} = \frac{l \cos θ}{(x + 10)^{2}}$ .......(2)
Equating luminous intensity from $(1)$ and $(2)$ , we get
$l = \frac{E_{A} x^{2}}{\cos θ} = \frac{E_{B} (x + 10)^{2}}{\cos θ}$

$\Rightarrow 900 x^{2} = 400 (x + 10)^{2} \Rightarrow \frac{x}{x + 10} = \frac{2}{3}$
⇒ 3x = 2x + 20
$\Rightarrow$ x = 20 cm
The distance between the source and the area at the initial phase was 20 cm.

Page No 455:

Answer:

Given,
Distance of the source from the table-top (r) = 60 cm or 0.6 m
Let I_obe the intensity of illumination.
Illuminance directly below the source $(E_{A})$ is given by,
$E_{A} = \frac{I_{0}}{(0.6)^{2}}$
⇒ I₀ = 15 × (0.6)²
= 5.4 candela

Let E_B be the illuminance at a point 80 cm away from the initial point.
So, $E_{B} = \frac{I_{0} \cos θ}{{(O B)}^{2}}$
From the figure, we get
cos $θ$ = $\frac{0.6}{1}$
OB = 1 m
Substituting the respective values in the above formula, we get
E_B $= 5.4 (\frac{0.6}{1}) = 3.24 lux$
So, the illuminance at a point on the table-top 80 cm away from the first point is 3.24 lux.

Page No 455:

Answer:

If the area is rotated about the incident light by any angle, the illuminance will not change.

Page No 455:

Answer:

Let the height of the source be h and the luminous intensity in the normal direction be I₀.
So, illuminance on the book $(E)$ is given by,
$E = \frac{I_{0} \cos θ}{r^{2}}$
From the figure we get
Cosθ = $\frac{h}{r}$
On substituting the value of Cosθ, we get
$E = \frac{I_{0} h}{r^{3}}$
But r = $\sqrt{R^{2} + h^{2}}$
$∴$ E = $\frac{I_{0} h}{{(r^{2} + h^{2})}^{\frac{3}{2}}}$
For maximum illuminance, $\frac{d E}{d H} = 0$
$\frac{d E}{d H} = \frac{I_{0} [{(R^{2} + h^{2})}^{\frac{3}{2}} - \frac{3}{2} h \times {(R^{2} + h^{2})}^{\frac{1}{2}} \times 2 h]}{{(R^{2} + h^{2})}^{3}} = 0$
$\Rightarrow {(R^{2} + h^{2})}^{\frac{1}{2}} [R^{2} + h^{2} - 3 h^{2}] = 0 \Rightarrow R^{2} - 2 h^{2} = 0 \Rightarrow h = \frac{R}{\sqrt{2}}$

Page No 455:

Answer:

Let x be the distance moved.
Let the illuminance of the candle and the lamp at a distance 1 m be I_C and I_L respectively. According to question,

Page No 455:

Answer:

Let I₁ be the intensity when placed at a distance 80 cm and I₂ be the intensity when placed at a distance 20 cm apart from the screen.

Now,

$\frac{I_{1}}{I_{2}} = {(\frac{80}{20})}^{2} = 16$

According to the question, let the new distance between the lamp and the screen be x such that even after covering the lamp with a thin paper the intensities at the screen is balanced.

So,

$\frac{0.49 I_{1}}{I_{2}} = {(\frac{x}{20})}^{2} \Rightarrow 0.49 \times 16 \times 400 = x^{2} \Rightarrow x = 56 cm$

Thus, the lamp has to be moved by 80 cm - 56 cm = 24 cm.

Page No 455:

Answer:

Total intensity of the 8 cd and the 12 cd light source is 20 cd.
Let d be the distance of the 80 cd source.
∴ Illuminance due to the 20 cd source $(E_{1})$ is:
$E_{1} = \frac{20}{{(0.4)}^{2}}$ .....(1)
∴ Illuminance due to the 80 cd source $(E_{2})$ is:
$E_{2} = \frac{80}{d^{2}}$ .....(2)
As, E₁ = E₂
$∴$ $\frac{20}{{(0.4)}^{2}}$ = $\frac{80}{d^{2}}$
⇒ d = 0.8 m = 80 cm.

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