JEE Physics 2025 Question with Solution

Given: A spherical refracting surface separates air and glass with refractive indices $n_1 = 1$ and $n_2 = 1.5$. The radius of curvature is $R$, and the condition is $PO = PI$.

Find: The distance $PO$.

Use refraction at a spherical surface:

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

From the given condition $PO = PI$, take equal magnitudes of object and image distances from point $P$, so $u = -v$.

Substitute $n_1 = 1$, $n_2 = 1.5$, and $u = -v$:

$$\frac{1.5}{v} - \frac{1}{-v} = \frac{1.5 - 1}{R}$$ $$\frac{1.5 + 1}{v} = \frac{0.5}{R}$$ $$\frac{2.5}{v} = \frac{0.5}{R}$$

Cross-multiplying,

$$2.5R = 0.5v$$ $$v = 5R$$

Since $PO = PI$ and the common distance equals $v$ in magnitude, therefore the distance is $5R$. The correct option is A.

Given: $n_1 = 1$ for air, $n_2 = 1.5$ for glass, and $PO = PI = x$.

Find: The value of $x = PO$.

Using the same spherical refraction formula,

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

Let $u = x$ and $v = -x$ in magnitude/sign convention form used in the extracted solution. Then,

$$\frac{1.5}{-x} - \frac{1}{x} = \frac{0.5}{R}$$ $$-\frac{2.5}{x} = \frac{0.5}{R}$$ $$x = -\frac{2.5R}{0.5} = -5R$$

As $x$ is a distance, take its magnitude:

$$x = 5R$$

Hence, the distance $PO$ is $5R$.