Physics Q42 – JEE-MAIN 2025

The radius of curvature for a thin convex lens are

10

cm and

15

cm respectively. The focal length of the lens is

12

cm. The refractive index of the lens material is

Clue 1 of 3: Lens Properties

Understand the characteristics of a convex lens, including its focal length and the sign convention for radii of curvature.

🧠 Full Solution Path

Step 1: Identify Given Values and Formula✦ Active

The given focal length of the convex lens is $f = + 12$ cm. The radii of curvature are $R_{1} = 10$ cm and $R_{2} = 15$ cm. For a biconvex lens, applying the Cartesian sign convention, the first surface (where light enters) has its center of curvature on the right, so $R_{1} = + 10$ cm. The second surface has its center of curvature on the left, so $R_{2} = - 15$ cm. We use the Lens Maker's Formula to find the refractive index $n$ :

\frac{1}{f} = (n - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})

Step 2: Substitute Values into the Formula○ Expand

Substitute the given values into the Lens Maker's Formula:

\frac{1}{12} = (n - 1) (\frac{1}{10} - \frac{1}{- 15})

\frac{1}{12} = (n - 1) (\frac{1}{10} + \frac{1}{15})

\frac{1}{12} = (n - 1) (\frac{3}{30} + \frac{2}{30})

\frac{1}{12} = (n - 1) (\frac{5}{30})

\frac{1}{12} = (n - 1) (\frac{1}{6})

Step 3: Solve for the Refractive Index○ Expand

Now, solve for $(n - 1)$ and then for $n$ :

n - 1 = \frac{1}{12} \times 6

n - 1 = \frac{6}{12}

n - 1 = 0.5

n = 0.5 + 1

n = 1.5

The refractive index of the lens material is $1.5$ .

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Physics Question 42 – JEE-MAIN 2025

Clue 1 of 3: Lens Properties

🧠 Full Solution Path