The critical angle for a denser-rarer interface is $45^{\circ}$. The speed of light in rarer medium is $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. The speed of light in the denser medium is:

<p>To find the speed of light in the denser medium, we can use Snell&#39;s Law at the critical angle, where the angle of refraction is $90^{\circ}$. Snell&#39;s Law states:</p> <p>$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$</p> <p>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.</p> <p>In this case, we have:</p> <p>$n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}$</p> <p>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:</p> <p>$n = \frac{c}{v}$</p> <p>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. </p> <p>For the rarer medium, we have:</p> <p>$n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1$</p> <p>Now we can rewrite Snell&#39;s Law as:</p> <p>$n_1 \sin{45^{\circ}} = 1 \cdot 1$</p> <p>$n_1 \sin{45^{\circ}} = 1$</p> <p>$n_1 = \frac{1}{\sin{45^{\circ}}}$</p> <p>Since $\sin{45^{\circ}} = \frac{1}{\sqrt{2}}$, we have:</p> <p>$n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$</p> <p>Now, we can find the speed of light in the denser medium using the formula for the index of refraction:</p> <p>$v_1 = \frac{c}{n_1}$</p> <p>$v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}$</p> <p>$v_1 = 2.12 \times 10^8 \mathrm{~m/s}$</p> <p>So, the speed of light in the denser medium is $2.12 \times 10^8 \mathrm{~m/s}$.</p>