IISER Physics - Optics
Exam Duration: 45 Mins Total Questions : 30
A standard 48 Cd lamp placed 36 cm from the screen of a photometer produces the same illumination there as a lamp of unknown intensity located 45 cm away. What is the luminous intensitty of the latter lamp?
- (a)
60 Cd
- (b)
75 Cd
- (c)
80 Cd
- (d)
100 Cd
Time of exposure for a photographic print is 10 second, when a lamp of 50 Cd is placed at 1m from it. Then another lamp of luminous intensity I is used, and is kept at 2m from it. If the time of exposure now is 20 s, the value of I (in Cd) is
- (a)
100
- (b)
25
- (c)
200
- (d)
20
If E is illuminance, t is time of exposure,
Et=constant or \(\phi\)
E2t2=E1t1
If I is luminous intensity, E=\(\frac{1}{r^{2}}\)
\(\therefore \frac{I_2}{(2)^{2}}\times t_2=\frac{I_t}{r_{1}^{2}}\times t_1\)
\(\frac{I}{(2)^{2}}\times20=\frac{50}{{I}^{2}}\times10\)
I=100 Cd
If an object is placed asymmetrically between two plane mirrors, inclined at an angle of 600, then the total number of images formed is
- (a)
5
- (b)
4
- (c)
2
- (d)
infinite
n=\(\frac{360^{0}}{\theta }\)-1=\(\frac{360^{0}}{60^{0}}\)-1=6-1=5
A concave mirror of focal length f(in air) is immersed in water (\(\mu \)=4/3). The focal length of the mirror in water will be
- (a)
f
- (b)
(4/3)f
- (c)
(3/4)f
- (d)
(7/3)f
Focal length of concave mirror is not changed by the medium in which it is placed.
A thin lens has focal length f and its aperture has radius r. It forms an image of intensity I. Now the central part of the aperture upto radius r/2 is blocked by an opaque material. Then the focal length and the intensity of the image will be
- (a)
\(\frac { f }{ 2 } and\frac { I }{ 2 } \)
- (b)
\( { f }{ } and\frac { I }{ 4} \)
- (c)
\(\frac {3 f }{ 4 } and\frac { I }{ 2 } \)
- (d)
\( { f }{ } and\frac { 3I }{ 2 } \)
Focal length of lens does not depend on aperture. If aperture changes, 'f' remains same and intensity chnages.Initial area of lens, A1=\(\pi r{2}\)
Blocked area of lens = \(\pi (\frac{r}{2})^{2}=\frac{\pi r^{2}}{4}\)
Transparent area of lens, \(A_2=\pi r^{2}-\frac{\pi r^{2}}{4}=\frac{3\pi r^{2}}{4}\)
\(\frac{A_2}{A_1}=\frac{3}{4}\)
Intensity ∝ Area
\(\frac{I_2}{I_1}=\frac{A_2}{A_1}=\frac{3}{4}; I_2=\frac{3}{4}I\)
The focal length of the convex lens depends upon
- (a)
frquency of the light ray
- (b)
wavelength of the light ray
- (c)
both of (a) and (b)
- (d)
NONE OF THESE
Focal length depends upon \(\mu\) which depends upon \(\lambda\).\(\mu\) does not depend upon frequency.
The distance between an object and the screen is 100 cm. A lens produces an image on the screen when placed at either of the positions 40 cm apart. The power of the lens is
- (a)
3D
- (b)
5D
- (c)
7D
- (d)
9D
\(f=\frac{D^{2}-d^{2}}{4D}\)
Given D=100 cm, d=40 cm
\(f=\frac{(100)^{2}-(40)^{2}}{4\times100}\)=\(\frac{(100-40)(100+40)}{4\times100}\)
=\(\frac{60\times140}{4\times100}=21\)cm
p=\(\frac{1}{f(in\quad m)}=\frac{1}{0.21}=\frac{100}{21}\)=5 D
An air bubble in a glass slab (\(\mu \)=1.5) is 5 cm deep when viewed from one face and 2 cm deep when viewed from one face and 2 cm deep when viewed from the opposite face. The thickness of the slab is
- (a)
10.5 cm
- (b)
7 cm
- (c)
10 cm
- (d)
7.5 cm
\(\mu=\frac{Read\quad Depth}{Apparent\quad Depth}\)=\(\frac{x}{y_1+y_2}=\frac{x}{5+2}; \)x=\(7\mu =7\times1.5 =10.5\)
How does refractive index \(\mu _{ }\) of a material vary with respect to wavelength \(\lambda \) ? A and B are constants
- (a)
\(\mu =A+\frac { B }{ { \lambda }^{ 2 } } \)
- (b)
\(\mu =A+B{ \lambda }^{ 2 }\)
- (c)
\(\mu =A+\frac { B }{ { \lambda }^{ } } \)
- (d)
\(\mu =A+B{ \lambda }^{ }\)
Cauchy's formula is
\(\mu=A+\frac{B}{\lambda^{2}}+\frac{C}{\lambda^{4}}+... \mu\cong A+\frac{B}{\lambda^{2}}\)
A ray is incident at angle of incidence i on one surface of a prism of small angle A and emerges normally from opposite surface. If the refractive index of the material of prism is \(\mu \), the angle of incidence i is nearly equal to
- (a)
A/2\(\mu \)
- (b)
A/2\(\mu \)
- (c)
\(\mu \)A
- (d)
\(\mu \)A/2
The retracting angle of a prism A is small. The correct statement for the dispersive power of a prism is that dispersive power.
- (a)
depends upon the material of prism
- (b)
depends upon both material and angle of prism
- (c)
depends only upon refracting angle of prism
- (d)
in small for all colours of white light
Dispersive power of the material of prism is
w=\(\frac{\mu_V-\mu_R}{\mu-1}\)
w depends upon the material of prism.
Green light of wavelength 5460 \(\mathring { A } \) is incident on an air-glass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be (c=3X108ms-1)
- (a)
3640\(\mathring { A } \)
- (b)
5460\(\mathring { A } \)
- (c)
4861\(\mathring { A } \)
- (d)
NONE OF THESE
\(\mu=\frac{c}{\vartheta }=\frac{\upsilon \lambda_a}{\upsilon \lambda_m}=\frac{\lambda_a}{\lambda_m}\)
\(\mu \lambda_m = \lambda_a\)where \(\lambda _a\) is wavelength in air and \(\lambda _m\) is wavelength is glass (medium).
or \(\mu \lambda\) = constant
\(\mu_1 \lambda_1=\mu_2 \lambda_2\)
For air \(\mu_1\)=1, \(\mu_2\)=1.5, =5460 \(\overset { O }{ A } \)
\(\lambda_2=\frac{\mu_1}{\mu_2} \lambda_1 = \frac{1}{1.5}\times 5460 \overset { O }{ A } = 3640 \overset { O }{ A } \)
A parallel beam of white light falls on a convex lens. Images of blue, yellow and red light are formed on the other side of the lens at a distance 20 cm, 20.5 cm and 21.4 cm respectively. The dispersive power of the material of the lens will be
- (a)
\(\frac { 6.19 }{ 1000 } \)
- (b)
\(\frac { 9 }{ 200 } \)
- (c)
\(\frac {14 }{ 205 } \)
- (d)
\(\frac { 5 }{ 214 } \)
Dispersive power, \(\omega = \frac{f_R-f_B}{f_Y}\)
=\(\frac{21.4-20.0}{20.5}=\frac{14}{205}\)
A planet is observed by an astronomical refracting telescope having an objective of focal length 16 m and eyepiece of focal length 2 cm.
A. The distance between objective and eyepiece is 16.02 m
B. The angular magnification is 800
C. The image of the planet is inverted
D. The objective is larger than eyepiece
- (a)
if A and B are correct
- (b)
if C and D are correct
- (c)
if B, C and D are correct
- (d)
if all are correct
Distance between objective and eyepiece = length of telescope = F0+Fe
Angular magnification = \(\frac{F_0}{F_e}=\frac{16 m }{0.02 m}=800\)
All the following statements are correct except
- (a)
The total focal length of an astronomical telescope is the sum of the focal lengths of its two lenses.
- (b)
The image formed by the astronomical telescope is always erect, because the effect of the combination of the two lenses is divergent
- (c)
The magnification of an astronomical telescope can be increased by decreasing the focal length of the eyepiece
- (d)
The magnifying power of the refracting type of astronomical telescope is the ratio of the focal length of the objective to that of the eyepiece
The image formed by the stronomical telescope is always inverted with respect to the object.
For the normal setting of a telescope
- (a)
only the object is at inifinity
- (b)
only the final image is at infinity
- (c)
both the object and the final image are at infinity
- (d)
neither the object nor the final images has to be infinity
Telescope is used to observe distant objects. So, object is already at infinity. For normal setting of telescope. final image is also at infinity.
While viewing a distant object with a telescope, suddenly a housefly sits on objective lens. The correct statement is that
- (a)
housefly will be seen enlarged in image
- (b)
housefly will be seen reduced in image
- (c)
intensity of image will be decreased
- (d)
intensity of image will be increased
A tow photons will be blocked by housefly. So, number of photons forming the image is reduced, i.e., the intensity of image will be decreased.
The resolving power of a telescope depends on
- (a)
focal length of eye lens
- (b)
focal length of objective lens
- (c)
length of the telescope
- (d)
diameter of the objective lens
Resolving power of telescope = \(\frac{D}{1.22 \lambda}\)
where D is diameter of objective lens and \(\lambda\) is wavelength of monochromatic incident light.
A person uses spectacle of power +2D. He is suffering from
- (a)
short sightedness or myopia
- (b)
long sightedness
- (c)
presbyopia
- (d)
asigmatism
As power of lens is positive, the person is wearing convex glasses. The person is suffering from long sightedness or hypermetropia.
Large apertures of the telescope are used for
- (a)
greater amplification
- (b)
greater resolution
- (c)
reducing lens aberration
- (d)
ease in manufacture
Resolving power of telescope = \(\frac{d}{1.22 \lambda}\)
where d is diameter (or aperture) of objective and \(\lambda\) is wavelength of incident light.
A dentist has a small mirror of focal length 16 mm. He views the cavity in the tooth of a patient by holding the mirror at a distance of 8 mm from the cavity. The magnification is
- (a)
1
- (b)
1.5
- (c)
2
- (d)
3
As, \(m=\frac { f }{ f-u } =\frac { -16 }{ -16-(-8) } =\frac { -16 }{ -18 } =2\)
With a concave mirror, an object is placed at a distance X1 from the principal focus, on the principal axis. The image is formed at a distance X2 from the principal focus. The focal length of the mirror is
- (a)
\({ x }_{ 1 }{ x }_{ 2 }\)
- (b)
\(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } \)
- (c)
\(\sqrt { \frac { { x }_{ 1 } }{ { x }_{ 2 } } } \)
- (d)
\(\sqrt { { x }_{ 1 }{ x }_{ 2 } } \)
As, \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
\(\therefore \frac{1}{(f-x_1)}+\frac{1}{(f-x_2)}=\frac{1}{f}\)
\(\Rightarrow \frac{f-x_2+f-x_1}{(f-x_1)(f-x_2)}=\frac{1}{f}\)
f2+fx2-fx1+x1x2=2f2-f(x1+x2)
f2=x1x2 ⇒ f=\(\sqrt{x_1x_2}\)
A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with \(\mu =1.65\). The focal length of the lens is nearly
- (a)
30 cm
- (b)
31 cm
- (c)
40 cm
- (d)
41 cm
As, \(\frac { 1 }{ f } =\left( 1.65-1 \right) \left( \frac { 2 }{ 40 } \right) \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 0.65 }{ 20 } \Rightarrow f=\frac { 20 }{ 0.65 } =30.77cm=31cm\)
A lens formed a virtual image 4 cm away from it when an object is placed 10 cm away from it. Which lens is this and what is its focal length?
- (a)
concave, 6.67 cm
- (b)
concave, 2.86 cm
- (c)
convex, 2.86 cm
- (d)
may be concave or convex, 6.67 cm
As, lens formula is \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -4 } -\frac { 1 }{ -10 } \Rightarrow \frac { 1 }{ f } =\frac { 1 }{ 10 } -\frac { 1 }{ 4 } \)7
\(\Rightarrow \frac { 1 }{ f } =\frac { 2-5 }{ 20 } =\frac { -3 }{ 20 } \Rightarrow f=-\frac { 20 }{ 3 } =-6.67cm\)
A lens of refractive index n is put in a liquid of refractive index \({ n }^{ \prime }\). If focal length of lens in air is f, then its focal length in liquid will be
- (a)
\(\frac { f{ n }^{ \prime }\left( n-1 \right) }{ { n }^{ \prime }-n } \)
- (b)
\(\frac { f\left( { n }^{ \prime }-n \right) }{ { n }^{ \prime }\left( n-1 \right) } \)
- (c)
\(\frac { { n }^{ \prime }\left( n-1 \right) }{ f\left( { n }^{ \prime }-n \right) } \)
- (d)
\(\frac { f{ n }^{ \prime }n }{ n-{ n }^{ \prime } } \)
By lens maker is formula, \(\frac{1}{f}=(\frac{n}{1}-1)\)\((\frac{1}{R_1}-\frac{1}{R_2})\) --- (i)
\(\frac{1}{f^{'}}=(\frac{n}{n^{'}}-1)(\frac{1}{R_1}-\frac{1}{R_2})\) --- (ii)
On dividing Eq. (ii) by Eq.(i), we get
\(\frac{f^{'}}{f}=\frac{(n-1)n^{'}}{n-n^{'}} \Rightarrow f^{'}=\frac{fn^{'}(n-1)}{n^{'}-n}\)
For a prism, its refractive index is \(\cot { \frac { A }{ 2 } } \) . Its minimum angle of deviation is
- (a)
180° - A
- (b)
180° - 2A
- (c)
90° - A
- (d)
\(\frac{A}{2}\)
As, \(\mu=\frac{sin(\frac{A+\delta_m}{2})}{sin \frac{A}{2}} \Rightarrow cot \frac{A}{2}=\frac{sin(\frac{A+\delta_m}{2})}{sin\frac{A}{2}}\)
\(\Rightarrow \frac{cos\frac{A}{2}}{sin \frac{A}{2}}=\frac{sin(\frac{A+\delta_m}{2})}{sin \frac{A}{2}}\)
\(\Rightarrow sin(\frac{\pi}{2}-\frac{A}{2})=sin(\frac{A+\delta_m}{2})\)
\(\Rightarrow \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta_m}{2} \Rightarrow \frac{\pi}{2}-A=\frac{\delta_m}{2}\)
\(\frac{\pi-2A}{2}=\frac{\delta_m}{2}\)
\(\therefore \delta_m=180^{0}-2A\)
An infinity long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is at distance u > f from the mirror. Its image will have a length of
- (a)
\(\frac { { f }^{ 2 } }{ u-f } \)
- (b)
\(\frac { uf }{ u-f } \)
- (c)
\(\frac { { f }^{ 2 } }{ u+f } \)
- (d)
\(\frac { uf }{ u+f } \)
From the reaction,
\(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}; \frac{1}{v}-\frac{1}{u}=\frac{1}{-f}\)
\(\frac{1}{v}=\frac{1}{u}-\frac{1}{f}\)
or \(v^{'}=(\frac{uf}{f-u})\)
∵ u>f, v is negative or |v|=\((\frac{uf}{u-f})>f\)
The end which is at infinity will have its image at focus.
ஃ Length if image L=|v|-f=\(\frac{f^{2}}{u-f}\)
A jar of height h is filled with a transparent liquid of refractive index \(\mu \) (figure). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the centre, the dot is invisible.
- (a)
\(d=\frac { h }{ \sqrt { { u }^{ 2 }-1 } } \)
- (b)
\(d=\frac { 2h }{ \left( { \mu }^{ 2 }-1 \right) } \)
- (c)
\(d=\frac { 2h }{ \sqrt { \left( { \mu }^{ 2 }-1 \right) } } \)
- (d)
None of the above
From Snell's law,
\(\sin { i } =\frac { 1 }{ \mu } \Rightarrow \tan { i } =\frac { d/2 }{ h } \) [from Eq.(i)]
\(d=\frac { 2h }{ \sqrt { { \mu }^{ 2 }-1 } } \quad \quad \left( \tan { i=\frac { 1 }{ \sqrt { { \mu }^{ 2 }-1 } } } \right) \)
Match the terms related to spherical mirror given in column I with their meaning given in column II and select the correct option from the choices given below
Column I | Column II |
A. Radius of curvature | 1. The distance between the pole and radius of curvature |
B. Principal axis | 2. The line joining the pole and the centre of curvature |
C. Focal length | 3. The distance between the pole and the centre of curvature |
- (a)
A B C 3 2 1 - (b)
A B C 1 2 3 - (c)
A B C 3 1 2 - (d)
A B C 2 1 3
A thin convex lens made from crown glass \(\left( \mu =\frac { 3 }{ 2 } \right) \)has focal length f. When it is measured in two different liquids having refractive indices \(\frac { 4 }{ 3 } and\quad \frac { 5 }{ 3 } \).It has the focal lengths f1 and f2 respectively. The correct relation between the focal length is
- (a)
f1 = f2 < f
- (b)
f1 > f and f2 becomes negative
- (c)
f2 > f and f1 becomes negative
- (d)
f1 and f2 both become negative
It is based on lens maker's formula and its magnification
i.e., \(\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
According to lens maker's formula, when the lens in the air
then, \(\frac{1}{f}=(\frac{3}{2}-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
\(\frac{1}{f} =\frac{1}{2x}\)
⇒ f=2x
Here, \((\frac{1}{x}=\frac{1}{R_1}-\frac{1}{R_2})\)
In case liquid , where refractive index is \(\frac{4}{3}\) and \(\frac{5}{3}\) , we get
Focal length in first liquid
\(\frac{1}{f_1}=(\frac{\mu_s}{\mu/s}-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
\(\Rightarrow \frac{1}{f_1}=(\frac{\frac{3}{2}}{\frac{4}{3}}-1)\frac{1}{x}\)
⇒ f1 is positive
\(\frac{1}{f_1}=\frac{1}{8x}=\frac{1}{4(2x)}=\frac{1}{4f}\Rightarrow f_1=4f\)
Focal length in second liquid,
\(\frac{1}{f_2}=(\frac{\mu_s}{\mu/s}-1)(\frac{1}{R_1}-\frac{1}{R_2})\)
\(\Rightarrow \frac{1}{f_2}=(\frac{\frac{3}{2}}{\frac{5}{3}}-1)(\frac{1}{x})\)
f2 is negative.