Chapter 10: Wave Optics (Optics)

Frequency: $\nu =\frac{c}{\lambda }=\frac{3\times 10^8}{589\times 10^{-9}}\approx 5.09\times 10^{14}$ Hz.

(a) Reflected light stays in air: wavelength 589 nm, frequency $5.09\times 10^{14}$ Hz, speed $3\times 10^8$ m/s.

(b) Refracted light: the frequency does not change ($5.09\times 10^{14}$ Hz).

Speed $v=\frac{c}{n}=\frac{3\times 10^8}{1.33}\approx 2.26\times 10^8$ m/s; wavelength ${\lambda }^{\prime }=\frac{\lambda }{n}=\frac{589}{1.33}\approx 443$ nm.

(a) 589 nm, $5.09\times 10^{14}$ Hz, $3\times 10^8$ m/s. (b) 443 nm, $5.09\times 10^{14}$ Hz, $2.26\times 10^8$ m/s.

• (a) Spherical: the light spreads equally in all directions from the point.
• (b) Plane: the lens turns the rays into a parallel beam.
• (c) Plane: the star is so far away that the small piece of its huge spherical wavefront reaching the Earth is practically flat.

(a) Spherical. (b) Plane. (c) Plane.

(a) $v=\frac{c}{n}=\frac{3.0\times 10^8}{1.5}=2.0\times 10^8$ m/s.

(b) No. The refractive index of glass depends on the wavelength (dispersion). It is larger for violet than for red, so violet light travels slower in glass. (That is why violet is bent the most by a prism.)

(a) $2.0\times 10^8$ m/s. (b) No; violet travels slower than red.

$x_4=\frac{4\lambda D}{d}$, so

$$\lambda =\frac{x_{4}d}{4D}=\frac{1.2\times 10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}=6.0\times 10^{-7}\text{m}=600\text{nm}$$

λ = 600 nm.

$I=4I_0{\cos }^2\frac{\varphi }{2}$, with $\varphi =\frac{2\pi }{\lambda }\times$ path difference.

• Path difference λ: $\varphi =2\pi$, ${\cos }^2\pi =1$, so $K=4I_0$.
• Path difference $\frac{\lambda }{3}$: $\varphi =\frac{2\pi }{3}$, ${\cos }^2\frac{\pi }{3}=\frac{1}{4}$, so $I=4I_0\times \frac{1}{4}=\frac{K}{4}$.

K/4.

The book does not give $D$ and $d$, so the answers are in terms of $\frac{D}{d}$. (The NCERT answer key uses $D=1.2$ m and $d=2$ mm, so $\frac{D}{d}=600$.)

(a) $x_3=\frac{3\lambda D}{d}=3\times 650\times 10^{-9}\times \frac{D}{d}=1.95\times 10^{-6}\frac{D}{d}$ m. With $\frac{D}{d}=600$: $x_3\approx 1.17$ mm.

(b) Bright fringes coincide where $n_1\times 650=n_2\times 520$, i.e. $\frac{n_1}{n_2}=\frac{520}{650}=\frac{4}{5}$. The first such place has $n_1=4$ (and $n_2=5$):

$$x=\frac{4\times 650\times 10^{-9}D}{d}=2.6\times 10^{-6}\frac{D}{d}\text{m}$$

With $\frac{D}{d}=600$: $x\approx 1.56$ mm.

(a) $1.95\times 10^{-6}D/d$ m (1.17 mm for D = 1.2 m, d = 2 mm). (b) $2.6\times 10^{-6}D/d$ m (1.56 mm), where the 4th bright fringe of 650 nm meets the 5th of 520 nm.