Q#1
The one parameter that determines the brightness of a light source sensed by an eye is(a) the energy of light entering the eye per second
(b) the wavelength of the light
(c) total radiant flux entering the eye
(d) total luminous flux entering the eye
Answer: (d)
The luminous flux is a quantity directly representing the total brightness producing capacity of a source.
Q#2
Three light sources A, B, and C emit equal amount of radiant energy per unit time. The wavelengths emitted by the three sources are 450 nm, 555 nm, and 700 nm respectively. The brightnesses sensed by an eye for the sources are Xₐ, Xᵦ, and X₍ respectively. Then,
(a) Xₐ > Xᵦ, X₍ > Xᵦ
(b) Xₐ > Xᵦ, Xᵦ > X₍
(c) Xᵦ > Xₐ, Xᵦ > X₍
(d) Xᵦ > Xₐ, X₍ > Xᵦ
Answer: (c)
The luminosity is maximum for the wavelength around 555 nm and falls off on both sides. Hence Xᵦ will be greater than the other two.
Q#3
As the wavelength is increased from violet to red, the luminosity
(a) continuously increases
(b) continuously decreases
(c) increases then decreases
(d) decreases then increases
Answer: (c)
As the wavelength is increased from violet (Approx 400 nm) the brightness increases up to wavelength around 555 nm and then decreases till the wavelength of red light (Approx 700 nm).
Q#4
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
(a) 10 lux
(b) 14 lux
(c) 20 lux
(c) 28 lux
Answer: (b)
The distance of the other point from the bulb, x = √(1² + 1²) = √2. Since the illuminance at a point is inversely proportional to the square of the distance from the source, the illuminance at the other point
= {1²/(√2)²}(40 lux)
= 40/2 = 20 lux.
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
(a) 10 lux
(b) 14 lux
(c) 20 lux
(c) 28 lux
Answer: (b)
The distance of the other point from the bulb, x = √(1² + 1²) = √2. Since the illuminance at a point is inversely proportional to the square of the distance from the source, the illuminance at the other point
= {1²/(√2)²}(40 lux)
= 40/2 = 20 lux.
Q#5
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
(a) 0.5%
(b) 1%
(c) 2%
(d) 4%
Answer: (c)
Suppose the original separation = 1 unit. When increased by 1%, the separation is 1.01 unit. If the illuminance at original separation = 100 lux, then the illuminance at increased separation = {1²/1.01²}(100 lux) = 98.03 lux.
Now the difference = 100 – 98.03 = 1.97 lux.
Hence the percentage decrease in illuminance = 1.97(100/100) = 1.97% ≈ 2%.
Q#6
A battery-operated torch is adjusted to send an almost parallel beam of light. It produces an illuminance 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
(a) 40 lux
(b) 20 lux
(c) 10 lux
(d) 5 lux.
Answer: (a)
Since the light beam is almost parallel, the luminous flux is the same per unit area at both the places. Since the illuminance is luminous flux incident per unit area, hence the illuminance at both the places will be the same.
Q#7
The intensity produced by a long cylindrical light source at small distance r from the source is proportional to
(a) 1/r²
(b) 1/r³
(c) 1/r
(d) none of these
Answer: (c)
Suppose the total luminous flux of the long cylindrical light source = F. The cylindrical area at a small distance r from the source on which it is incident = 2πrL, {where L is the length of the source, leaving the end circular areas which are negligible}.
Hence the luminous intensity, I = F/2πrL
Since F and L are constant,
I ∝ 1/r.
Q#8
A photographic plate placed at 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
(a) 3 s
(b) 12 s
(c) 24 s
(d) 48 s.
Answer: (b)
Suppose the photographic plate is receiving P luminous flux per unit area (Illuminance) from the source when it is at 5 cm from the source. Here total exposure received = P*3 = 3P units.
At 10 cm the illuminance = P(5²/10²) = P/4
{Since the illuminance is inversely proportional to the square of the distance from the source}
If 't' is the time needed for the same exposure, then
t(P/4) = 3P
t = 3 x 4 = 12 s.
Q#9
A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12 s to get a good exposure. If the source is rotated by 60° about one of its diameters, the time needed to get the same exposure will be
(a) 6 s
(b) 12 s
(c) 24 s
(d) 48 s.
Answer: (c)
Let the illuminance in the first case =P. In 12 s it gets exposure of 12 x P = 12P units.
Since the illuminance is directly proportional to Cosθ, where θ is the angle made by the normal to the area with the direction of incident radiation, the illuminance after rotation = P*cosθ = P*cos60° = P/2. Now if the time for same exposure be, t, then
t(P/2) = 12P
t = 2 x 12 = 24 s
Q#10
A point source of light moves in a straight line parallels to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at a portion varies with its distance r from the source as
(a) I ∝ 1/r
(b) I ∝ 1/r²
(c) I ∝ 1/r³
(d) I ∝ 1/r⁴.
Answer: (c)
Let us draw a diagram.
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
(a) 0.5%
(b) 1%
(c) 2%
(d) 4%
Answer: (c)
Suppose the original separation = 1 unit. When increased by 1%, the separation is 1.01 unit. If the illuminance at original separation = 100 lux, then the illuminance at increased separation = {1²/1.01²}(100 lux) = 98.03 lux.
Now the difference = 100 – 98.03 = 1.97 lux.
Hence the percentage decrease in illuminance = 1.97(100/100) = 1.97% ≈ 2%.
Q#6
A battery-operated torch is adjusted to send an almost parallel beam of light. It produces an illuminance 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
(a) 40 lux
(b) 20 lux
(c) 10 lux
(d) 5 lux.
Answer: (a)
Since the light beam is almost parallel, the luminous flux is the same per unit area at both the places. Since the illuminance is luminous flux incident per unit area, hence the illuminance at both the places will be the same.
Q#7
The intensity produced by a long cylindrical light source at small distance r from the source is proportional to
(a) 1/r²
(b) 1/r³
(c) 1/r
(d) none of these
Answer: (c)
Suppose the total luminous flux of the long cylindrical light source = F. The cylindrical area at a small distance r from the source on which it is incident = 2πrL, {where L is the length of the source, leaving the end circular areas which are negligible}.
Hence the luminous intensity, I = F/2πrL
Since F and L are constant,
I ∝ 1/r.
Q#8
A photographic plate placed at 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
(a) 3 s
(b) 12 s
(c) 24 s
(d) 48 s.
Answer: (b)
Suppose the photographic plate is receiving P luminous flux per unit area (Illuminance) from the source when it is at 5 cm from the source. Here total exposure received = P*3 = 3P units.
At 10 cm the illuminance = P(5²/10²) = P/4
{Since the illuminance is inversely proportional to the square of the distance from the source}
If 't' is the time needed for the same exposure, then
t(P/4) = 3P
t = 3 x 4 = 12 s.
Q#9
A photographic plate is placed directly in front of a small diffused source in the shape of a circular disc. It takes 12 s to get a good exposure. If the source is rotated by 60° about one of its diameters, the time needed to get the same exposure will be
(a) 6 s
(b) 12 s
(c) 24 s
(d) 48 s.
Answer: (c)
Let the illuminance in the first case =P. In 12 s it gets exposure of 12 x P = 12P units.
Since the illuminance is directly proportional to Cosθ, where θ is the angle made by the normal to the area with the direction of incident radiation, the illuminance after rotation = P*cosθ = P*cos60° = P/2. Now if the time for same exposure be, t, then
t(P/2) = 12P
t = 2 x 12 = 24 s
Q#10
A point source of light moves in a straight line parallels to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at a portion varies with its distance r from the source as
(a) I ∝ 1/r
(b) I ∝ 1/r²
(c) I ∝ 1/r³
(d) I ∝ 1/r⁴.
Answer: (c)
Let us draw a diagram.
d = distance between the parallel lines of the table and the line of movement of the source.
θ is the angle between normal and the direction of propagation of the light.
If I = Luminous intensity in the direction of propagation SP, and
SL = perpendicular from S to the line of the table.
The illuminance at a point is
E = Icosθ/r²
= Icos(90°-α)/r²
= Isinα/r²
= I(d/r)/r², {From the right angled triangle SPL}
= Id/r³
Here I and d are fixed, hence Id = constant. Thus
E ∝ 1/r³
Q#11
Figure (22-Q1) shows a glowing mercury tube. The intensities at point A, B and C are related as
(a) B > C > A
(b) A > C > B
(c) B = C > A
(d) B = C < A.
Answer: (d)
Points C and B are equidistant and symmetrically located, hence the intensities at these points are equal. Compare point A and any of the points C or B. The minimum distance of C from the glowing mercury tube is longer than the minimum distance of A from the tube. Also, the angle subtended by the tube at C is narrower than the angle subtended at A. Hence the intensity at A will be greater than B or C. Thus the option (d).
θ is the angle between normal and the direction of propagation of the light.
If I = Luminous intensity in the direction of propagation SP, and
SL = perpendicular from S to the line of the table.
The illuminance at a point is
E = Icosθ/r²
= Icos(90°-α)/r²
= Isinα/r²
= I(d/r)/r², {From the right angled triangle SPL}
= Id/r³
Here I and d are fixed, hence Id = constant. Thus
E ∝ 1/r³
Q#11
Figure (22-Q1) shows a glowing mercury tube. The intensities at point A, B and C are related as
(a) B > C > A
(b) A > C > B
(c) B = C > A
(d) B = C < A.
Points C and B are equidistant and symmetrically located, hence the intensities at these points are equal. Compare point A and any of the points C or B. The minimum distance of C from the glowing mercury tube is longer than the minimum distance of A from the tube. Also, the angle subtended by the tube at C is narrower than the angle subtended at A. Hence the intensity at A will be greater than B or C. Thus the option (d).
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