JEE Physics 2025 Question with Solution

Given: A bi-convex lens has both radii equal in magnitude to $\frac{1}{6} \, \text{cm}$.

Find: A new convex lens with $R_1 \neq R_2$ but the same power.

Use the lens-maker's formula:

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

For the original bi-convex lens, taking the sign convention,

$$R_1=\frac{1}{6} \, \text{cm}, \qquad R_2=-\frac{1}{6} \, \text{cm}$$

So,

$$\frac{1}{R_1}-\frac{1}{R_2}=6-(-6)=12$$

To keep the same power, the new lens must satisfy

$$\frac{1}{R_1}-\frac{1}{R_2}=12$$

Now check the possible combination $\frac{1}{5} \, \text{cm}$ and $\frac{1}{7} \, \text{cm}$ with convex-lens sign convention as magnitudes:

$$R_1=\frac{1}{5} \, \text{cm}, \qquad R_2=-\frac{1}{7} \, \text{cm}$$

Then,

$$\frac{1}{R_1}-\frac{1}{R_2}=5-(-7)=12$$

Hence the power remains unchanged.

Therefore, the correct option is A.

The solution labels the correct option as B, but based on the listed options provided here, the pair $\frac{1}{5} \, \text{cm}$ and $\frac{1}{7} \, \text{cm}$ corresponds to option A.

Given: Same refractive index material and same power are required.

Principle: For lenses made of the same material in air, equal power means the quantity $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ must remain the same.

For the original symmetric bi-convex lens,

$$(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=(\mu-1)\left(\frac{2}{R}\right)$$

With $R=\frac{1}{6} \, \text{cm}$,

$$\frac{2}{R}=\frac{2}{1/6}=12$$

So the new lens must satisfy

$$\frac{1}{R_1}+\frac{1}{|R_2|}=12$$

Testing the first available pair of magnitudes:

$$\frac{1}{1/5}+\frac{1}{1/7}=5+7=12$$

So this pair gives the same power.

Therefore, the required combination is $\frac{1}{5} \, \text{cm}$ and $\frac{1}{7} \, \text{cm}$, which is option A in the provided list.