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 Bihar Board Class 12th Physics Solutions Chapter 10 Wave Optics

Bihar Board Class 12 Physics Wave Optics Textbook Questions and Answers

Question 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? The Refractive index of water is 1.33.
Answer:
λ = Wavelength of moncochromatic light
= 589 nm, c – speed of light
= 3 x 108 ms-1.

(a) For reflected light
λ’ = wavelength of reflected light
v = frequency of reflected light = ?
v = speed of reflected light = ?

(i) As the wavelength of the reflected light remains unchanged, so λ’ = λ = 589 nm = 589 x 10-9 m.

(ii) As the reflection takes place in the same medium, so
v = speed of light = C = 3 x 108 ms-1.

(iii) Using relation, C = vλ, we get
v=Cλ=3×108589×10−9=5.093×1014 Hz

(B) For refracted light—Let λw be the wavelength of light in water
vw = frequency of light in water = ?
uw = speed of light in water = ?

Using the relation,

In moving from one medium to another medium, the frequency of light does not change as the colour of light does not changes, so

Question 2.
What is the shape of the wavefront 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 wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) When the light diverging from a point source, then the wavefront is of a converging spherical shape.
(b) When the point source is placed on the principal focus of the convex lens, then the rays of light emerging from the lens are parallel to each other, so the wave front will be plane.
(c) In this case the shape of the wavefront is almost plane because the source of light i.e., the star is far away from earth, thus a small area on the surface of a large sphere is nearly plane.

Question 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 x 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?
Answer:
Here, μ = R.I. of glass = 1.5.
C = 3.0 x 108 ms-1.
ug = speed of light in glass = ?
Using the relation, μ = Cv, we get
vg=Cμg=3×1081.5=2×108 ms−1

(b) No, the speed of light in glass is not independent of the colour of light. According to Cauchy’s formula, the refractive index depends on the colour i.e., wavelength of light as :

as C, a, b are constants.

∴ v α λ2. Thus clearly the speed of light is directly proportional to the square of the wavelength of colour of the light.

Also we know that λv < λr i.e., wavelength of violet colour is lesser than the wavlength of red colour, thus clearly violet colour travels slower in glass than the red component.

Question 4.
In a Young’s double-slit experiment, the slits are spearated 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.
Answer:
Here, d = 0.28 mm = 0.28 x 10-3 m.
D = Distance between slits and screen = 1.4 m
n = order of fringe = 4
y = 1.2 cm = 1.2 x 10-2 m
λ = ?
For constructive interference,
y=nλDd, we get 

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength X, the intensity of light at a point on the screen where path difference is X, is K units. What is the intensity of light at a point where path difference is λ3?
Answer:
Here, I = K when path difference = λ
I’ = ? When path difference = λ3
We know that the intensity I is given by –
I = 2I0 (1 + cos Φ) ……………………………………………………… (1)
When Φ = phase difference
When path difference is λ, let Φ be the phase difference.
∴ From relation,

Let Φ, be the phase difference for a path difference λ3.

Question 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 wavelength coincide?
Answer:
The distance between the slits is 2 mm and the distance between the plane of slits and screen is 120 cm.

Here, λ1 = 650 nm = 650 x 10-9 m = 6500 x 10-10 m.
λ2 = 520 nm = 520 x 10-9 m = 5200 x 10-10 m.

(a) n = 3, λ = λ1 = 6500 x 10-10 m.
D = 120 cm = 1.20 m, d = 2 mm = 2 x 10-3 m
x = distance of third bright fringe from central, maximum = ?

Using the relation,

(b) Let n be the least number of fringes of λ1 (= 650 nm) which coincide with (n + 1) fringes of λ2 (= 520 nm),

Question 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 43.
Answer:
Here, X = 600 nm = 600 x 10-9 m
= 6000 x 10-10 m.
θ = angular separation = 0,2′
D = distance of the screen = 1 m

Question 8.
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5).
Answer:
Here, μg = R.I of glass = 1.5.
Let p Brewster angle = ?
Also we know that for air to glass transition,
μ = tan p
∴ tan p = 1.5
or p = tan-1 1.5
or p = 56.3°.

Question 10.9.
Light of wavelength 5000 A 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?
Answer:
Here, λ = wavelength of light = 5000 A° = 5000 x 10-10 m.
Let λ’, V be the wavelength and frequency of the reflected light.
c = Velocity of light in vacuum
= 3 x 108 ms-1.
If v be the frequency of the incident light, then
v=cλ=3×1085000×10−10=6×10710−7
= 6 x 1014 Hz.
The wavelength, p and speed of the reflected light remains same, so it frequency (v’) also remains same in all media.

Here XY is a plane reflecting surface.
OA and AB are the incident and reflected ray.
i = angle of incidence
r = angle of reflection As AB ⊥ OA,
∴ i + r = 90°
Also, according to laws of reflection,
r = i
∴ i + i = 90°
or 2 i = 90°
∴ i=902=45∘

Question 10.
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer:
Here,
λ = wavelength of light = 400 nm
= 400 x 10-9 m
a = sizeofaparture=4mm
= 4 x 10-3 m
Zf = Fresnel’s distance
= distance for which the ray optics is good approximation =?

Using the relation,
Zf=a2λ, we get 
Zf=(4×10−3)2400×10−9 m=40 m