QJEEM23Ph0b18: JEE Main 2023 Physics - Class 12 JEEM23Ph0b18 Solution: The critical angle for a denser-rarer interface is . The speed…

To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is $90^{\circ}$ . Snell's Law states:

$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$

where $n_1$ and $n_2$ are the indices of refraction for the denser and rarer media, respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

In this case, we have:

$n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}$

The critical angle is given as $45^{\circ}$ , and the speed of light in the rarer medium is given as $3 \times 10^8 \mathrm{~m/s}$ . We can find the index of refraction of the rarer medium using the formula:

$n = \frac{c}{v}$

where $c$ is the speed of light in a vacuum ( $3 \times 10^8 \mathrm{~m/s}$ ), and $v$ is the speed of light in the medium.

For the rarer medium, we have:

$n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1$

Now we can rewrite Snell's Law as:

$n_1 \sin{45^{\circ}} = 1 \cdot 1$

$n_1 \sin{45^{\circ}} = 1$

$n_1 = \frac{1}{\sin{45^{\circ}}}$

Since $\sin{45^{\circ}} = \frac{1}{\sqrt{2}}$ , we have:

$n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

Now, we can find the speed of light in the denser medium using the formula for the index of refraction:

$v_1 = \frac{c}{n_1}$

$v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}$

$v_1 = 2.12 \times 10^8 \mathrm{~m/s}$

So, the speed of light in the denser medium is $2.12 \times 10^8 \mathrm{~m/s}$ .