CBSE Class 12 Physics Chapter-10 Important Questions – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 12 Physics Chapter 10 – Wave Optics 3 Marks Questions prepared by expert Physics teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 12 Physics Chapter-10 – Wave Optics
CBSE Class 12 Physics Important Questions Chapter 10 – Wave Optics
3 Marks Questions
1. State Brewster law? Using this law prove that, at the polarizing angle of incidence, the reflected and transmitted rays are perpendicular to each other?
Ans. according to Brewster law the longest of the angle of polarization for transparent medium is equal to the refractive index of the medium.
Proof. Using Snell’s law
———(1)
——-(2)
2. In a single slit experiment, how is the angular width of central bright fringe maximum changed when
1) The slit width increased
2) The distance between the slit and the screen is increased.
3) Light of smaller wavelength is used.
Ans. In single slit diffraction
(a) When slit width‘d’ is increased. 
decreases
(b) When ‘D’ is increased, width of central bright fringe will become maximum i.e
increase.
(c) When light of smaller wavelength is used, the width of central bright maximum
decrease.
3. In a young’s double slit experiment, the slit are repeated at 0.24mm. The screen is 1.2m away from the slits. The fringe width is 0.3cm calculate the wavelength of light used in the experiment?
Ans. 
4. Two coherent sources whose intensity ratio is 81:1 produce interference fringes. Calculate the ratio of intensity of maxima and minima in the interference pattern?
Ans 
Intensity
5. Using Huygens’s principle deduce the laws of refraction?
Ans. According to Huygens’s theory each point on AB given rise to new wave fronts give taken by the wavelets to reach from
——(1)
Substituting in equation (1)
Since time is independent of equation
Term containing AO must be zero.
Hence proved Snell’s law
6. A young’s double slit experiment using light of wavelength 400 nm, interference fringes of width to 600nm, and the separation between the slits is halved. If one wants the observed fringe width on the screen to be the same in the two cases,find the ratio of the distance between the screen and the plane of the interfering
Ans. Let 
be the distance between the screen and the sources, when light of wavelength 400nm is used.
In order to obtain the same fringe width
——-(2)
From equation (1) and (2)
sources in the two arrangements.
7. In young’s double slit experiment while using a source of light of wavelength
, the fringe width obtained is 0.6cm. If the distance between the slit and the screen is reduced to half, calculate the new fringe width?
Ans. 
——-(1)
New fringe width
8. What is polarization of light? What type of waves show the property of polarization? Name any two methods to produce plane polarized light
Ans. The phenomenon of restricting the vibrations of a light vector in a particular direction in a plane perpendicular to the direction of propagation of light is called polarisation of light. Transverse waves show the property of polarisaiton.
Two methods to produce plane polarised light
(1)Polarisation by Reflection
(2) Polarization by scattering
9. Draw the curve depicting, variation of intensity in the interference pattern in young’s double slit experiment. State conditions for obtaining sustained interference of light?
Ans.
Conditions for sustained interference of light
(1) Two sources must be coherent sources of light.
(2)Two sources should exist light waves continuously. Intensity monochromatic
10. What is the shape of the wave front in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wave front of light from a distant star intercepted by the Earth.
Ans. (a) The shape of the wave front in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.
(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.
11. In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Ans. Distance between the slits, d = 0.28 mm =
m
Distance between the slits and the screen, D = 1.4 m
Distance between the central fringe and the fourth (n = 4) fringe,
u = 1.2 cm =
m
In case of a constructive interference, we have the relation for the distance between the two fringes as:
Where,
n = Order of fringes = 4
= Wavelength of light used
∴
Hence, the wavelength of the light is 600 nm.
12. In a double-slit experiment the angular width of a fringe is found to be 
on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.
Ans. Distance of the screen from the slits, D = 1 m
Wavelength of light used, 
Angular width of the fringe in 
Angular width of the fringe in water =
Refractive index of water,
Refractive index is related to angular width as:
Therefore, the angular width of the fringe in water will reduce to 
13. Light of wavelength 5000 
falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Ans. Wavelength of incident light, = 5000 ¦ = 
m
Speed of light, 
Frequency of incident light is given by the relation,
The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 ¦ and its frequency is
.
When reflected ray is normal to incident ray, the sum of the angle of incidence, 
and angle of reflection, 
is 
.
According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:
Therefore, the angle of incidence for the given condition is
.
14. The 6563 
line emitted by hydrogen in a star is found to be red shifted by 15 ¦. Estimate the speed with which the star is receding from the Earth.
Ans. Wavelength of line emitted by hydrogen,
= 6563 
¦
=
Star’s red-shift, 
Speed of light, 
Let the velocity of the star receding away from the Earth be v.
The red shift is related with velocity as:
Therefore, the speed with which the star is receding away from the Earth is 
15. You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Ans. Let an object at O be placed in front of a plane mirror MO’ at a distance r (as shown in the given figure).
A circle is drawn from the centre (O) such that it just touches the plane mirror at point O’. According to Huygens’ Principle, XY is the wavefront of incident light.
If the mirror is absent, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r (as shown in the given figure).
can be considered as a virtual reflected ray for the plane mirror. Hence, a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the object distance (r).
16. A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Ans. Wavelength of light beam, = 500 nm = 
Distance of the screen from the slit, D = 1 m
For first minima, n = 1
Distance between the slits = d
Distance of the first minimum from the centre of the screen can be obtained as:
x = 2.5 mm = 
It is related to the order of minima as:
Therefore, the width of the slits is 0.2 mm.
