NCERT Solutions for Class 12 Physics, Chapter10: Wave Optics are available here for download. You can download these solutions with the help of download link given at the end of this article. Wave optics is one of the most important chapters of Class 12 NCERT textbook. Questions from this chapter of NCERT textbook are frequently asked in CBSE Class 12 board exams and other competitive exams like NEET, JEE Main etc. Students must understand these solutions to score well in exams.
NCERT Solutions for Class 12 Physics ‒ Chapter 10: Wave Optics are given below:
Question 10.1: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) Refracted light? Refractive index of water is 1.33.
Given: = 589 nm,
n = 1.33
(a) For Reflected light all parameters will be same. So, = 589 nm
⇒ c = 3 × 108 m / s
⇒ f = c / = 5.09 × 1014 Hz.
(b) For refracted light, wavelength and speed will be different but frequency will be same.
New speed, v = (c) / n = (3 × 108)/(1.33) = 2.26 × 108 m/s.
New wavelength, ’ = / μ = 444 nm.
Question 10.2: 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
Question 10.3: (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s–1)
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
(a) New speed, v = c / n = (3 × 108) / 1.5 = 2 × 108 m / s.
(b) No, it is not independent. Violet will travel slower than red.
Question 10.4: In a Young’s double-slit experiment, the slits are separated by0.28 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.
d = 0.28 mm
D = 1.4 m
Fringe spacing = 1.2 cm
D (/ d) = 1.2 x 10‒2
Calculation yields, = 600 nm.
Question 10.5: In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?
I1 = I2 = I
Phase difference = ф
Resultant intensity is given by, IR = I1 + I2 + 2 (I1 I2)1/2 cos ф
When path difference = λ, phase difference ф =
IR = I + I + 2 √(I)2 cos0o = 4I = K
When path difference = λ/3, phase difference ф =2π/3
IR’ = I + I + 2 √(I)2 cos (2π/3) = 2I + 2I (‒1/2) = K/4.
Question 10.6: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Let, D be the distance of screen from two slits and d be the distance between two slits.
λ1 = 650 nm = 650 × 10‒9 m, λ2 = 520 nm = 520 × 10‒9 m
For 3rd bright fringe, n = 3.
x = n λ1 (D/d) = 3 × 650 [(D/d)] nm
nth fringe due to λ2 coincides with (n ‒ 1)th fringe due to λ1.
n λ2 = (n ‒ 1) λ1 ⇒ n = 5.
Least distance required, x = n λ2 (D/d) = 5 × 520 (D/d) = 2600 D/d nm.
The values of D and d are not given in the question.
Question 10.7: In a double-slit experiment the angular width of a fringe is found to be 0.2° 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.
D = 1 m
Angular width = 0.2
λ = 600 nm
n = 4 / 3
Angular separation = 0.2 = λ / d
Which gives, d = 3 micro meters
Using, n = 4 / 3
New speed, v = c / n = 2.25 × 108 m / s
f = c / λ = 5 × 1014 Hz
Now, λ’ = v / f = 450 nm
β’ = λ’D / d = 0.15.
Question 10.8: What is the Brewster angle for air to glass transition?
(Refractive index of glass = 1.5.)
Brewster angle, n = tan (ip)
⇒ tan-1(1.5) = 56.31o
Question 10.9: Light of wavelength 5000 Å 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?
λ = 5000 angstroms
Wavelength and frequency will be same for the reflected light.
Frequency = c / λ = (3 × 108)/(5000 × 10‒10) = 6 × 1014 Hz
Reflected ray will be normal to the incident ray for angle of incidence = 45o.
Question 10.10: Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
a = 4 mm
Using the formula, ZF = a2 / λ = (4×10‒3)2/(4 × 10‒7) = 40 m.
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