JEE Physics 2025 Question with Solution

Given: The power of the original lens is $4\,\text{D}$.

Find: The power of one part when the symmetric thin biconvex lens is cut into four equal parts by planes $AB$ and $CD$.

When the lens is divided symmetrically, the focal length of each part becomes double of the original lens. Since power is inversely proportional to focal length,

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

so if focal length doubles, power becomes half.

Therefore,

$$P_{\text{part}} = \frac{P_{\text{original}}}{2} = \frac{4\,\text{D}}{2} = 2\,\text{D}$$

Hence, the power of each part is $2\,\text{D}$.

Therefore, the correct option is C.

Given: Original lens power is $4\,\text{D}$.

Find: Power of one of the four equal parts.

Use the lens power relation:

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

When the symmetric lens is cut into equal parts as shown, each part has doubled focal length compared with the original lens.

So,

$$f_{\text{part}} = 2f_{\text{original}}$$

Hence,

$$P_{\text{part}} = \frac{1}{f_{\text{part}}} = \frac{1}{2f_{\text{original}}} = \frac{1}{2}P_{\text{original}}$$

Substituting $P_{\text{original}} = 4\,\text{D}$,

$$P_{\text{part}} = \frac{1}{2} \times 4\,\text{D} = 2\,\text{D}$$

Thus, the power of one part is $2\,\text{D}$.