A ray of light AO in vacuum is incident on a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Snell's law,$n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (vacuum),$i = 60^o$,$n_2 = \mu$ (glass),and $r = 30^o$.$1 \cdot \sin 60^o = \mu

Vedclass · Accepted Answer

$2a + 2b$

Vedclass · Answer

(D) According to Snell's law,$n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (vacuum),$i = 60^o$,$n_2 = \mu$ (glass),and $r = 30^o$.$1 \cdot \sin 60^o = \mu \cdot \sin 30^o$$\frac{\sqrt{3}}{2} = \mu \cdot \frac{1}{2} \Rightarrow \mu = \sqrt{3}$.The optical path length is defined as the sum of the products of refractive index and geometric path length for each medium.Optical path length $= n_{vacuum} \cdot AO + n_{glass} \cdot OB$.From the geometry of the figure:$AO = \frac{a}{\cos 60^o} = \frac{a}{1/2} = 2a$.$OB = \frac{b}{\cos 30^o} = \frac{b}{\sqrt{3}/2} = \frac{2b}{\sqrt{3}}$.Optical path length $= 1 \cdot (2a) + \sqrt{3} \cdot \left( \frac{2b}{\sqrt{3}} \right) = 2a + 2b$.

$A$ ray of light $AO$ in vacuum is incident on a glass slab at an angle of $60^o$ and refracted at an angle of $30^o$ along $OB$ as shown in the figure. The optical path length of the light ray from $A$ to $B$ is

Similar Questions

Light travels through water in a beaker. The height of the water column is $h$. If the refractive index of water is $\mu_{w}$ and $C$ is the velocity of light in air,the time taken by light to travel through the water will be:

$A$ beam of monochromatic blue light of wavelength $4200 Å$ in air travels in water of refractive index $4/3$. Its wavelength in water will be: (in $Å$)

State Snell's law of refraction.

The refractive index of water is $1.33$. The direction in which a man under water should look to see the setting sun is

Vedclass Test Series

Exam Paper Generator

Online Exam Module