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?
Answer:
Wavelength of the light beam, λ1 = 650 nm
Wavelength of another light beam, λ2 = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
(i) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,
x = \(n\;\lambda_{1}\;(\frac{D}{d})\)
For third bright fringe, n = 3
Therefore \(x = 3\times650\frac{D}{d}= 1950\frac{D}{d}\)
(b) Let, the nth bright fringe due to wavelength λ2 and (n–1)th bright fringe due to wavelength λ2 coincide on the screen. The value of n can be obtained by equating the conditions for bright fringes:
nλ2 = (n−1)λ1
520n = 650n – 650
650 = 130n
Therefore n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
\(x = n\;\lambda_{2}\;\frac{D}{d} = 5\times 520\frac{D}{d}=2600\frac{D}{d}\,nm\)
Note: The value of d and D are not given in the question.
7. In a double-slit experiment, 0.2° is found to be the angular width of a fringe 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 \(\frac{4}{3}\).
Answer:
Distance of the screen from the slits, D = 1m
Wavelength of light used, λ1 = 600 nm
Angular width of the fringe in air θ1 = 0.2°
Angular width of the fringe in water = θ2
Refractive index of water, μ = \(\frac43\)
\(\mu=\frac{\theta_{1}}{\theta_{2}}\) is the relation between the refractive index and the angular width
\(\theta_{2}=\frac{3}{4}\theta_{1}\)
\(=\frac{3}{4}\times 0.2=0.15\)
Therefore, 0.15° is the reduction in the angular width of the fringe in water.
8. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer:
Refractive index of glass, μ = 1.5
Consider Brewster angle = θ
Following is the relation between the Brewster angle and the refractive index:
tan θ = μ
θ = tan−1(1.5) = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°
9. Light of wavelength 5000 Armstrong 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:
Wavelength of incident light, [\lambda][λ] = 5000 Armstrong = 5000 x 10-10 m
Speed of light, c =3 x 108 m
Following is the relation for the frequency of incident light:
\(v = \frac{c}{\lambda}= \frac{3\;\times \;10^{8}}{5000\;\times \;10^{-10}} = 6 \times10^{14}\)
The wavelength and frequency of incident light is equal to the reflected ray.
Therefore, 5000 Armstrong and 6 × 1014Hz is the wavelength and frequency of the reflected light.
When reflected ray is normal to incident ray, the sum of the angle of incidence, ∠i and angle of reflection, ∠r is 90°.
From laws of reflection we know that the angle of incidence is always equal to the angle of reflection
\(\angle i+\angle r = 90 ^\circ\)
i.e. \(\angle i+\angle i = 90 ^\circ\)
Hence, \(\angle i=\frac{90}{2} = 45^\circ\)
Therefore, 45° is the angle of incidence.
10. Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.
Answer:
Fresnel’s distance (ZF) is the distance which is used in ray optics for a good approximation. Following is the relation,
\(Z_{F}=\frac{a^{2}}{\lambda}\)
Where,
Aperture width, a = 4 mm = 4 × 10-3 m
Wavelength of light, λ = 400 nm = 400 × 10-9 m
\(Z_{F}=\frac{(4\times10^{-3})^{2}}{400\times10^{-9}}= 40\,m\)
Therefore, 40m is the distance for which the ray optics is a good approximation.
11. The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.
Answer:
λ = 6563 Å
Δλ = 15 Å
Since the star is receding, the velocity (v) is negative.
Δλ = – vλ/c
v = – cΔλ/λ
= – (3 x 108) x (15 Å/ 6563 Å)
= – 6.86 x 105 m/s
12. Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?
Answer:
According to Newton’s Corpuscular theory, velocity of light in the denser medium (water) is greater than the velocity of light in the rarer medium (vacuum). This was experimentally wrong.
At the angle of incidence (i) of the light of velocity v, the angle of refraction is r.
Due to the change in the medium, the change in the velocity of light in water is v
Using Snells law,
c sin i = v sin r ......(1)
The relation between the velocities and the refractive index is
v/c = μ ........(2)
v/c = sin i/sin r = μ ......(3)
But μ > 1 so v > c is not possible
Huygens wave theory is consistent with the experiment.
13. 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.
Answer:
Consider an object O placed in front of the plane mirror MO’ at a distance r. The object is taken as the centre point O and a circle is drawn such that it just touches the plane mirror at point O’.
According to Huygens’ Principle, XY is the wavefront of the incident light. If the mirror was not present, then a similar wavefront X’Y’ (as XY) would form behind O’ at distance r. X’Y’ can be considered as a virtual reflected ray for the plane mirror. Therefore, 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).
