Q: Three light sources A, B and C emit equal amount of radiant energy per unit time. The wavelengths emitted by the three source are 450 nm, 555 nm and 700 nm respectively. The brightness sensed by an eye for the sources areXA,XB andXC respectively. Then, ________ .

(c) XB>XA,XB>XCWavelength of light B is 555 nm. It has the highest luminosity; hence, XB will be highest. Again, 450 nm is nearer to 555 nm than 700 nm.∴ 555-450=105But 700−555=145So, XA's brightness will be greater than that of XC.

Question 1

What is the luminous flux of a source emitting radio waves?

Accepted Answer

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

Question 2

The luminous flux of a 1 W sodium vapour lamp is more than that of a 10 kW source of ultraviolet radiation. Comment.

Accepted 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.

Question 3

Light is incident normally on a small plane surface. If the surface is rotated by an angle of 30° about the incident light, does the illuminance of the surface increase, decreases or remain same? Does your answer change if the light did not fall normally on the surface?

Accepted Answer

If the surface is rotated by 30°, 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 30° rotation brings the table closer to the normal of the surface, the illuminance will increase; otherwise, it will decrease.

Question 4

A bulb is hanging over a table. At which portion of the table is the illuminance maximum? If a plane mirror is placed above the bulb facing the table, will the illuminance on the table increase?

Accepted 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.

Question 5

The sun is less bright at morning and evening as compared to at noon although its distance from the observer is almost the same Why?

Accepted 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.

Question 6

Why is the luminous efficiency small for a filament bulb as compared to a mercury vapour lamp?

Accepted 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.

Question 7

The yellow colour has a greater luminous efficiency as compared to the other colours. Can we increase the illuminating power of a white light source by putting a yellow plastic paper around this source?

Accepted 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.

Question 8

The one parameter that determines the brightness of a light source sensed by an eye is ____________ .

Accepted Answer

(d) Total luminous flux entering the eye.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.

Question 9

Three light sources A, B and C emit equal amount of radiant energy per unit time. The wavelengths emitted by the three source are 450 nm, 555 nm and 700 nm respectively. The brightness sensed by an eye for the sources areXA​,XB​  andXC​  respectively. Then, ________ .

Accepted Answer

(c) XB​>XA​,XB​>XC​Wavelength of light B is 555 nm. It has the highest luminosity; hence, XB​ will be highest. Again, 450 nm is nearer to 555 nm than 700 nm.∴  555-450=105But 700−555=145So, XA​'s brightness will be greater than that of XC​.

Question 10

As the wavelength is increased from violet to red, the luminosity ____________ .

Accepted Answer

(c) Increases then decreases.The luminosity first increases up to 555 nm and then decreases.

Question 11

An electric bulb is hanging over a table at a height of 1 m above it. The illuminance on the table directly below the bulb is 40 lux. The illuminance at a point on the table 1 m away from the first point will be about ___________ .

Accepted Answer

(b) 14 luxHere,r=2​tanθ=ABBC​=1Io​=40 lux θ=tan−1(1)=450​ The illuminance is given byE=r2Io​cosθ​=(2​)240×cos(450)​=14 lux ​

Question 12

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1%, the illuminance will decrease (nearly) by ____________ .

Accepted Answer

(c) 2%.Illuminance is given by;E=r2Io​cosθ​θ=00rΔr​=1%E=r2Io​​​ Differentiating,dE=−2r3Io​​drAs approximation differentials are replaced by Δ,ΔE=−2r3Io​​Δr⇒ΔE=−2r2Io​​(rΔr​)⇒ΔE=−2E(rΔr​)⇒EΔE​=−2(rΔr​)⇒EΔE​=−2×1%=−2%​ Since, negative sign implies decrease; hence, illuminance decreases by 2%.

Question 13

A battery-operated torch is adjusted to send an almost parallel beam of light. It produces an illuminancle of 40 lux when the light falls on a wall 2 m away. The illuminance produced when it falls on a wall 4 m away is close to _________ .

Accepted Answer

(a) 40 lux.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.

Question 14

The intensity produced by a long cylindrical light source at a small distance r from the source is proportional to _________ .

Accepted Answer

(c) r1​.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=rdθdx′=r′dθ Now,dA=dxdy=rdθdydA′=dx′dy=r′dθdydA′dA​=r′r​⇒rdA​=r′dA′​=dω The luminous flux going through the solid angle dω  will be:dF=Idω Now,dF=IrdA​ If the surfaces are inclined at an angle α ,dF=IrdAcosα​ Now, illuminance is defined asE=dAdF​=IrdAcosα​⇒E∝r1​

Question 15

A photographic plate placed a distance of 5 cm from a weak point source is exposed for 3 s. If the plate is kept at a distance of 10 cm from the source, the time needed for the same exposure is _____________ .

Accepted Answer

(b) 12s.Here,d1​ =5 cm=0.05 md2​ =10 cm = 0.1 md2​ =10 cm = 0.1 mt2​ =?Let the actual incident illuminance be t1​ =3 sLet the iluminance at 3cm distance be t2​ =?Let the iluminance at 10cm distance be Eo​cosθ = 1Ed1​Now,Ed2​⇒  t1​  =Ed1​k​cosθ = 1⇒Eo​k​=253​Similarly,Ed1​ =d1​2Eo​​⇒ t2​ =253​× 102 =12 s

Question 16

A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12s to get a good exposure. If the source is rotated by 60° about one of its diameter, the time needed to get the same exposure will be ___________ .

Accepted Answer

(c) 24 s.Here,t1​θ1​θ2​t2​​=12s=00=600=?​ Let the distance be r.Let the incident luminosity be Eo​. We have,Eθ​1=r2Eo​cosθ1​​t1​αEθ​11​⇒t1​=Eo​cosθ1​r2k​⇒12=Eo​cos0r2k​⇒Eo​r2k​=12​ Similarly,t2​=Eo​cosθ2​r2k​=cos(600)12​⇒t2​=12×2=24s​

Question 17

A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at this portion varies with its distance r from the source as ___________ .

Accepted Answer

(c) 1∝r31​.Let the distance between the parallel straight lines be L.Angle with normal = ​θWe know,I=r2Io​cosθ​From the above figure, we  getI=r2Io​cos(900−α)​⇒I=r2Io​sinα​⇒I=r2Io​sinα​⇒I=r2Io​sinα​L =  constant for parallel moving sourceSo, ⇒I=r2Io​sinα​⇒I=r2Io​sinα​⇒I=r3Io​L​

Question 18

Figure shows a glowing mercury tube. The intensities at point A, B and C are related as __________ .

Accepted Answer

(d) B = C 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 0° with the source, intensity of A will be greater than C or B.

Question 19

The brightness-producing capacity of a source:

Accepted Answer

(c) Depends on its power(d) Depends on the wavelength emitted.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.

Question 20

A room is illuminated by an extended source. The illuminance at a particular portion of a wall can be increased by:

Accepted Answer

(a), (b), (c), (d) 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.

Chapter 22 : Photometry