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.
solution :
$$\lambda_{air} = 589 nm = 589 \times 10^{-9}m$$
$$_{a}{\mu}_{w} = 1.33$$
$$c = 3 \times 10^{8}m/s$$
(a) For reflected light . On reflection, no change occurs in wavelength, frequency and speed of the incident light.
Wavelength of reflected light ,
$$\lambda_{air} = 589 nm = 589 \times 10^{-9}m$$
$$f = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{589 \times 10^{-9}}$$
$$f = 5.09 \times 10^{14}Hz$$
Speed of reflected light ,
$$v = c= 3 \times 10^{8}m/s$$
(b) For refracted light ,
On refraction, the frequency f remains the same but both wavelength and speed get changed.
Wavelength of refracted light,
$$_{a}\mu_{w} = \frac{\lambda_{air}}{\lambda_{water}}$$
$${\lambda_{water}}= \frac{\lambda_{air}}{_{a}\mu_{w}}$$
$$\lambda_{water} = \frac{589 \times 10^{-9}}{1.33}$$
$$\lambda_{water} = 4.42 \times 10^{-7}m$$
Frequency of refracted light
$$f = 5.09 \times 10^{14}Hz$$
Speed of refracted light ,
$$_a\mu_w =\frac{c}{v_{water}}$$
$$v_{water} = \frac{c}{_a\mu_w}$$
$$v_{water} = \frac{3 \times 10^{8}}{1.33}$$
$$v_{water} = 2.66 \times 10^{8}m/s$$
10.2 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.
Solution :
(a) The geometrical shape of the wavefront will be diverging spherical wavefront (SWF). It is because the locus of the all points equidistant from the point source is a sphere.
(b) When a point source is placed at the focus of a convex lens the Ray of light emerging out of the lens are parallel. Therefore the geometrical shape of the wavefront will be plain wavefront (WF).
(c) As a spherical wave front of a star source of light advance it's curvature goes on decreasing ( radius increase) since the star is very far off at the infinity the wavefront intercepted by the earth is plane wavefront.
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 $3.0 \times 10^{8} m/s$) (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?
Solution :
(a) $$\mu = \frac{c}{v}$$
$$v = \frac{c}{\mu}$$
Speed of light in glass
$$v= \frac{3.0 \times 10^{8}}{1.5}$$
$$v = 2.0 \times 10^{8} m/s$$
(b) No
$$v =\frac{c}{\mu}$$
Since
$$_{1}\mu_{2} = \frac{\mu_{2}}{\mu_{1}}$$
$$_{1}\mu_{2}= \frac{v_{1}}{v_{2}}$$
$$_{1}\mu_{2}= \frac{\lambda_{1}}{\lambda_{2}}$$
$$\frac{v_{1}}{v_{2}} =\frac{\lambda_{1}}{\lambda_{2}}$$
$$v \propto \lambda$$
So
$$\lambda_{R} > \lambda_{v}$$
$$v_{R} > v_{v}$$
Speed of violet colour travels slower than Red colour.
10.4 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.
Solution:
$$d = 0.28mm = 0.28 \times 10^{-3}m$$
$$D = 1.4 m$$
$$x_{4} = 1.2 cm = 1.2 \times 10^{-2}m$$
$$n = 4$$
The distance of nth bright fringe from central bright fringe is
$$x_{n} = n \lambda \frac{D}{d}$$
$$x_{4} = 4 \lambda \frac{D}{d}$$
$$\lambda = \frac{x_{4}.d}{4D}$$
$$\lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$$
$$\lambda = 6 \times 10^{-7}m$$
$$\lambda = 600 \times 10^{-2} \times 10^{-7}m$$
$$\lambda = 600 \times 10^{-9}m$$
$$\lambda = 600nm$$
10.5 In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is K units. What is the intensity of light at a point where path difference is $\lambda/3$?
Solution :
Case 1.
$$\bigtriangleup x = \lambda$$
$$I = K$$
Case 2.
$$\bigtriangleup x = \frac{\lambda}{3}$$
$$I = ?$$
In Young's double slit experiment, if $\phi$ is the phase difference between two light waves , then resultant intensity is
$$I = I_{1} + I_{2} + 2(\sqrt{I_{1}I_{2}})cos\phi$$
$$I = I_{o} +I_{o} + 2(\sqrt{I_{o}.I_{o}}) cos\phi$$
$$I = 2.I_{o} + 2I_{o}cos\phi$$
$$I = 2.I_{o}(1+cos\phi)$$
$$I = 2.I_{o}(1+2cos^{2}\frac{\phi}{2}-1)$$
$$I =2.I_{o}(2cos^{2}\frac{\phi}{2})$$
$$I = 4I_{o}(cos^{2}\frac{\phi}{2})$$
$$\phi = \frac{2 \pi}{\lambda} \times \bigtriangleup x$$
(i) When path difference is $\lambda$, the phase difference $\phi$ is
$$\phi = \frac{2 \pi}{\lambda} \times \lambda$$
$$\phi = 2\pi$$
$$I = 4 \times I_{o}(cos\frac{2\pi}{2})^{2}$$
$$I = 4 \times I_{o}(cos\pi)^{2}$$
$$I = 4 \times I_{o}(-1)^{2}$$
$$I = 4 \times I_{o} = K$$
(ii) When path difference is $\frac{\lambda}{3}$, the phase difference is
$$\phi = \frac{2 \pi}{\lambda} \times \frac{\lambda}{3}$$
$$\phi = \frac{2 \pi}{3}$$
$$\phi = 120^{o}$$
$$I = 4I_{o}(cos^{2}\frac{120}{2})$$
$$I = 4I_{o}(cos^{2}60^{o})$$
$$I = 4I_{o}(cos^{2}60^{o})$$
$$I = 4I_{o}(\frac{1}{2})^{2})$$
$$I = 4I_{o} \times \frac{1}{4}$$
$$I = \frac{K}{4}$$
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?
Solution :
(a) distance of the nth bright fringe on the screen from the central maximum is
$$x_{n} = \frac{nD \lambda}{d}$$
$$x_{3} = \frac{3 \times D \times 650 \times 10^{-9}}{d}$$
$$x_{3} = 1.95 \times 10^{-4} \times \frac{D}{d}m$$
(b) Let least distance x from the central maximum
$$x = n_{1} \beta_{1} = n_{2} \beta_{2}$$
$$n_{1} \frac{D \lambda_{1}}{d} = n_{2} \frac{D \lambda_{2}}{d}$$
$$n_{1}.\lambda_{1} = n_{2}.\lambda_{2}$$
$$\frac{n_{1}}{n_{2}} = \frac{\lambda_{2}}{\lambda_{1}}$$
$$\frac{n_{1}}{n_{2}} = \frac{520nm}{650nm}$$
$$\frac{n_{1}}{n_{2}} = \frac{4}{5}$$
For least bright fringe
$$n_{1} = 4$$
$$n_{2} = 5$$
Required least distance,
$$x_{4} = \frac{4D \lambda_{1}}{d}$$
$$x_{4} = \frac{4D \times 650 \times 10^{-9}}{d}$$
$$x_{4} = 2.6 \times 10^{-6} \times \frac{D}{d}m$$