Physics Q37 – JEE-MAIN 2025

A bi-convex lens has radius of curvature of both the surfaces same as

\frac{1}{6}

cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (

R_{1} \neq R_{2}

), without any change in lens power then possible combination of

R_{1}

and

R_{2}

is:

Clue 1 of 3: Lens Power

The power of a lens depends on its refractive index and the radii of curvature of its surfaces.

🥷

Ninja Strategy•Condition Check and Direct Calculation

First, eliminate options that violate the explicit condition ( $R_{1} \neq R_{2}$ ). Then, calculate the required power factor from the original lens and directly check which remaining option satisfies it using the lens maker's formula.

🧠 Full Solution Path

Step 1: Determine the power factor for the original lens✦ Active

The original lens is bi-convex with radii of curvature of both surfaces having a magnitude of $R = \frac{1}{6}$ cm. For a bi-convex lens, using the Cartesian sign convention, if light travels from left to right, the first surface has $R_{1} = + R$ and the second surface has $R_{2} = - R$ . The power $P$ is given by the lens maker's formula:

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

Substituting the values for the original lens:

P_{o r i g i n a l} = (n - 1) (\frac{1}{1 / 6} - \frac{1}{- 1 / 6}) = (n - 1) (6 - (- 6)) = (n - 1) (12)

Thus, the power factor $(\frac{1}{R_{1}} - \frac{1}{R_{2}})$ for the original lens is $12$ .

Step 2: Apply the condition for the new lens○ Expand

The new convex lens must have the same power as the original lens, so its power factor must also be $12$ . The problem also states that the new lens must have different radii of curvatures on both sides ( $R_{1} \neq R_{2}$ ). For a convex lens, $R_{1}$ is positive and $R_{2}$ is negative. We need to find an option where $(\frac{1}{R_{1}^{'}} - \frac{1}{R_{2}^{'}}) = 12$ , with $R_{1}^{'}$ and $R_{2}^{'}$ being the magnitudes given in the options, and $R_{1}^{'} \neq R_{2}^{'}$ .

💡 Teacher's Secret Hint

Remember to use the correct sign convention for $R_{1}$ and $R_{2}$ for a convex lens.

Step 3: Evaluate the options○ Expand

Let's check each option by calculating its power factor $(\frac{1}{R_{1}^{'}} - \frac{1}{R_{2}^{'}})$ :

Option 1: $R_{1}^{'} = \frac{1}{6}$ cm, $R_{2}^{'} = - \frac{1}{9}$ cm. $\frac{1}{1 / 6} - \frac{1}{- 1 / 9} = 6 - (- 9) = 15 \neq 12$ .

Option 2: $R_{1}^{'} = \frac{1}{3}$ cm, $R_{2}^{'} = - \frac{1}{3}$ cm. $\frac{1}{1 / 3} - \frac{1}{- 1 / 3} = 3 - (- 3) = 6 \neq 12$ . Also, $R_{1}^{'} = R_{2}^{'}$ , which violates the condition $R_{1} \neq R_{2}$ .

Option 3: $R_{1}^{'} = \frac{1}{5}$ cm, $R_{2}^{'} = - \frac{1}{7}$ cm. $\frac{1}{1 / 5} - \frac{1}{- 1 / 7} = 5 - (- 7) = 12$ . This matches the required power factor and $R_{1}^{'} \neq R_{2}^{'}$ .

Option 4: $R_{1}^{'} = \frac{1}{3}$ cm, $R_{2}^{'} = - \frac{1}{7}$ cm. $\frac{1}{1 / 3} - \frac{1}{- 1 / 7} = 3 - (- 7) = 10 \neq 12$ .

Therefore, option 3 is the correct combination.

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Physics Question 37 – JEE-MAIN 2025

Clue 1 of 3: Lens Power

🧠 Full Solution Path