Chapter 10 Wave Optics Solutions
Question - 1 : - Monochromatic light of wavelength589 nm is incident from air on a water surface. What are the wavelength,frequency and speed of (a) reflected, and (b) refracted light? Refractive indexof water is 1.33.
Answer - 1 : -
Wavelength of incidentmonochromatic light,
λ = 589nm = 589 × 10−9 m
Speed oflight in air, c = 3 × 108 m/s
Refractiveindex of water, μ = 1.33
(a) The ray will reflect back in thesame medium as that of incident ray. Hence, the wavelength, speed, andfrequency of the reflected ray will be the same as that of the incident ray.
Frequencyof light is given by the relation,
Hence, the speed, frequency, andwavelength of the reflected light are 3 × 108 m/s,5.09 ×1014 Hz, and 589 nmrespectively.
(b) Frequency of light does not depend onthe property of the medium in which it is travelling. Hence, the frequency ofthe refracted ray in water will be equal to the frequency of the incident orreflected light in air.
Refracted frequency, ν = 5.09 ×1014 Hz
Speed oflight in water is related to the refractive index of water as:
Wavelength of light in water isgiven by the relation,
Hence, the speed, frequency, andwavelength of refracted light are 2.26 ×108 m/s,444.01nm, and 5.09 × 1014 Hzrespectively.
Question - 2 : - What is the shape of thewavefront in each of the following cases:
(a) Light diverging from a pointsource.
(b) Light emerging out of a convexlens when a point source is placed at its focus.
(c) The portion of thewavefront of light from a distant star intercepted by the Earth.
Answer - 2 : -
(a) The shapeof the wavefront 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) Theshape of the wavefront in case of a light emerging out of a convex lens when apoint source is placed at its focus is a parallel grid. This is shown in thegiven figure.
(c) Theportion of the wavefront of light from a distant star intercepted by the Earthis a plane.
Question - 3 : - (a) Therefractive index of glass is 1.5. What is the speed of light in glass? Speed oflight in vacuum is 3.0 × 108 m s−1)
(b) Is the speed of light in glassindependent of the colour of light? If not, which of the two colours red andviolet travels slower in a glass prism?
Answer - 3 : -
(a) Refractiveindex of glass, μ = 1.5
Speed oflight, c = 3 × 108 m/s
Speed oflight in glass is given by the relation,
Hence, the speed of light inglass is 2 × 108 m/s.
(b) The speed of light in glass isnot independent of the colour of light.
Therefractive index of a violet component of white light is greater than therefractive index of a red component. Hence, the speed of violet light is lessthan the speed of red light in glass. Hence, violet light travels slower thanred light in a glass prism.
Question - 4 : - In a Young’s double-slitexperiment, the slits are separated by 0.28 mm and the screen is placed 1.4 maway. The distance between the central bright fringe and the fourth brightfringe is measured to be 1.2 cm. Determine the wavelength of light used in theexperiment.
Answer - 4 : -
Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m
Distancebetween the slits and the screen, D =1.4 m
Distancebetween the central fringe and the fourth (n =4) fringe,
u = 1.2 cm = 1.2 × 10−2 m
In caseof a constructive interference, we have the relation for the distance betweenthe two fringes as:
Where,
n = Order of fringes = 4
λ =Wavelength of light used
∴
Hence, the wavelength of thelight is 600 nm.
Question - 5 : - In Young’s double-slit experimentusing monochromatic light of wavelengthλ, the intensity of light at a point onthe screen where path difference is λ, is K units.What is the intensity of light at a point where path difference is λ /3?
Answer - 5 : -
Let I1 and I2 be the intensity of the two light waves. Theirresultant intensities can be obtained as:
Where,
= Phase difference between thetwo waves
Formonochromatic light waves,
Phase difference = 
Since path difference = λ,
Phasedifference, 
Given,
I’ = K
When path difference
Phase difference, 
Hence, resultant intensity, 
Using equation (1), we can write:
Hence, the intensity of light at a point where thepath difference is 
is 
units.
Question - 6 : - A beam of light consisting of twowavelengths, 650 nm and 520 nm, is used to obtain interference fringes in aYoung’s double-slit experiment.
(a) Find the distance of the thirdbright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance fromthe central maximum where the bright fringes due to both the wavelengthscoincide?
Answer - 6 : - Wavelength of the light beam, 
Wavelength of another lightbeam, 
Distance of the slits from thescreen = D
Distancebetween the two slits = d
(a) Distance of the nth brightfringe on the screen from the central maximum is given by the relation,
(b) Letthe nth brightfringe due to wavelength 
and (n − 1)th bright fringe due towavelength 
coincide on the screen. Wecan equate the conditions for bright fringes as:
Hence, the least distance fromthe central maximum can be obtained by the relation:
In a double-slit experiment theangular width of a fringe is found to be 0.2° on a screen placed 1 m away. Thewavelength of light used is 600 nm. What will be the angular width of thefringe if the entire experimental apparatus is immersed in water? Takerefractive index of water to be 4/3.
Answer - 7 : -
Distance of the screen from theslits, D = 1 m
Wavelengthof light used, 
Angularwidth of the fringe in 
Angular width of the fringe inwater =
Refractiveindex of water,
Refractive index is related toangular width as:
Therefore, the angular width ofthe fringe in water will reduce to 0.15°.
What is the Brewster angle forair to glass transition? (Refractive index of glass = 1.5.)
Answer - 8 : - Refractive index of glass, 
Brewster angle = θ
Brewsterangle is related to refractive index as:
Therefore, the Brewster angle forair to glass transition is 56.31°.
Light of wavelength 5000 Å fallson a plane reflecting surface. What are the wavelength and frequency of thereflected light? For what angle of incidence is the reflected ray normal to theincident ray?
Answer - 9 : -
Wavelength of incident light, λ =5000 Å = 5000 × 10−10 m
Speed oflight, c = 3 × 108 m
Frequencyof incident light is given by the relation,The wavelength and frequency ofincident light is the same as that of reflected ray. Hence, the wavelength ofreflected light is 5000 Å and its frequency is 6 × 1014 Hz.
Whenreflected ray is normal to incident ray, the sum of the angle ofincidence, 
and angle ofreflection, 
is 90°.
Accordingto the law of reflection, the angle of incidence is always equal to the angleof reflection. Hence, we can write the sum as:
Therefore, the angle of incidencefor the given condition is 45°.
Estimate the distance for whichray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer - 10 : -
Fresnel’s distance (ZF)is the distance for which the ray optics is a good approximation. It is givenby the relation,
Where,
Aperturewidth, a = 4 mm = 4 ×10−3 m
Wavelengthof light, λ = 400 nm = 400 × 10−9 m
Therefore, the distance for whichthe ray optics is a good approximation is 40 m.