JEE PYQ: Wave Optics Question 13
Question 13 - 2020 (03 Sep Shift 2)
Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first wave travels a path $L_1$ through a medium of refractive index $n_1$ while the second wave travels a path of length $L_2$ through a medium of refractive index $n_2$. After this the phase difference between the two waves is :
(1) $\frac{2\pi}{\lambda}\left(\frac{L_2}{n_1} - \frac{L_1}{n_2}\right)$
(2) $\frac{2\pi}{\lambda}\left(\frac{L_1}{n_1} - \frac{L_2}{n_2}\right)$
(3) $\frac{2\pi}{\lambda}(n_1 L_1 - n_2 L_2)$
(4) $\frac{2\pi}{\lambda}(n_2 L_1 - n_1 L_2)$
Show Answer
Answer: (3)
Solution
The distance traversed by light in a medium of refractive index $\mu$ in time $t$ is given by $d = vt$ …(i)
where $v$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,
$\Delta = ct = c \times \frac{d}{v}$ [from equation (i)]
$= d \frac{c}{v} = \mu d$ …(ii) $(\because \mu = \frac{c}{v})$
This distance is the equivalent distance in vacuum and is called optical path.
Optical path for first ray which travels a path $L_1$ through a medium of refractive index $n_1 = n_1 L_1$
Optical path for second ray which travels a path $L_2$ through a medium of refractive index $n_2 = n_2 L_2$
Path difference $= n_1 L_1 - n_2 L_2$
Now, phase difference $= \frac{2\pi}{\lambda} \times$ path difference $= \frac{2\pi}{\lambda} \times (n_1 L_1 - n_2 L_2)$
BETA
