QJEEM24Ph4df6: JEE Main 2024 Physics - Class 12 JEEM24Ph4df6 Solution: For the thin convex lens, the radii of curvature are at and…

To find the refractive index of the material of a thin convex lens, we can make use of the Lensmaker's Formula. The Lensmaker's formula is given by:

$\frac{1}{f} = \left( \frac{\mu - 1}{\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

where

• $f$ is the focal length of the lens,
• $\mu$ is the refractive index of the material of the lens,
• $R_1$ is the radius of curvature of the first surface (convex side, positive),
• $R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).

Given in the question, the radii of curvature are $15 \, \mathrm{cm}$ and $30 \, \mathrm{cm}$ respectively, and the focal length $f = 20\, \mathrm{cm}$ .

It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention. For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \, \mathrm{cm}$ and $R_2 = -30 \, \mathrm{cm}$ .

Substituting the given values into the Lensmaker's Formula, we get:

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right)$

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right)$

$\frac{1}{20} = (\mu - 1) \left( \frac{2 + 1}{30} \right)$

$\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right)$

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right)$

$\frac{1}{20} = \frac{\mu - 1}{10}$

Now, solve for $\mu$ :

$\mu - 1 = \frac{10}{20}$

$\mu - 1 = 0.5$

$\mu = 1.5$

Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.