JEE Main 2020 — Ray Optics Question with Solution

The optical path between any two points is proportional to the time of travel.

The distance traversed by light in a medium of refractive index $\mathrm{\mu }$ in time $t$ is given by $\mathrm{d}=\mathrm{vt}$ .......(i)

Where $\mathrm{v}$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

$s=ct$

$=\mathrm{c}\times \frac{\mathrm{d}}{\mathrm{v}}$   [from equation (i)]

$=\mathrm{d}\frac{\mathrm{c}}{\mathrm{v}}=\mathrm{\mu d}$ .....(ii) (Since$\mathrm{\mu }=\frac{\mathrm{c}}{\mathrm{v}}$)

This distance is the equivalent distance in vacuum and is called optical path.

Here, optical path for first ray $={\mathrm{n}}_1{\mathrm{L}}_1$

Optical path for second ray $={\mathrm{n}}_2{\mathrm{L}}_2$

Path difference $={\mathrm{n}}_1{\mathrm{L}}_1-{\mathrm{n}}_2{\mathrm{L}}_2$

Now, phase difference $=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times$path difference

$=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}\times \left({\mathrm{n}}_1{\mathrm{L}}_1-{\mathrm{n}}_2{\mathrm{L}}_2\right)$