JEE Physics 2025 Question with Solution

Given: Two thin convex lenses have focal lengths $f_1 = 30 \, \text{cm} = 0.3 \, \text{m}$ and $f_2 = 10 \, \text{cm} = 0.1 \, \text{m}$. The separation is $d = 10 \, \text{cm} = 0.1 \, \text{m}$.

Find: The equivalent power of the combination.

First, convert the focal lengths into powers using

$$P = \frac{1}{f}$$

So,

$$P_1 = \frac{1}{0.3} \approx 3.33 \, \text{D}$$

and

$$P_2 = \frac{1}{0.1} = 10 \, \text{D}$$

For two lenses separated by a distance $d$, the equivalent power is

$$P_{\text{eq}} = P_1 + P_2 - d \, P_1 P_2$$

Substituting the values,

$$P_{\text{eq}} = 3.33 + 10 - (0.1 \times 3.33 \times 10)$$ $$P_{\text{eq}} = 13.33 - 3.33$$ $$P_{\text{eq}} = 10 \, \text{D}$$

Using exact values for verification,

$$P_1 = \frac{10}{3} \, \text{D}, \qquad P_2 = 10 \, \text{D}$$

Then,

$$P_{\text{eq}} = \frac{10}{3} + 10 - \left(0.1 \times \frac{10}{3} \times 10\right)$$ $$P_{\text{eq}} = \frac{10}{3} + \frac{30}{3} - \frac{10}{3}$$ $$P_{\text{eq}} = \frac{30}{3} = 10 \, \text{D}$$

Therefore, the power of the lens combination is $10 \, \text{D}$. The correct option is D.

Given: $f_1 = 30 \, \text{cm}$, $f_2 = 10 \, \text{cm}$, and separation $d = 10 \, \text{cm}$.

Find: The power of the combination.

Use the separated-lens formula directly:

$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$

Substituting in metres,

$$\frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)}$$

Simplifying,

$$\frac{1}{f_{\text{eq}}} = 10 \, \text{D}$$

Hence, the equivalent power is $10 \, \text{D}$. The correct option is D.

This shortcut works because power is the reciprocal of focal length, so once $\frac{1}{f_{\text{eq}}}$ is found, the answer is obtained immediately.