Physics Q42 – JEE-MAIN 2026

A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be

Clue 1 of 3: Lens Properties

Understand how the focal length of a lens changes when its geometry is altered.

🧠 Full Solution Path

Step 1: Determine the focal length of the original biconvex objective lens✦ Active

For a symmetric biconvex lens with radii of curvature $R_{1} = R$ and $R_{2} = - R$ , the focal length $f_{o}$ is given by the lens maker's formula:

\frac{1}{f_{o}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = (μ - 1) \frac{2}{R}

Thus, $f_{o} = \frac{R}{2 (μ - 1)}$ .

Step 2: Determine the focal length of the new plano-convex objective lens○ Expand

When the biconvex lens is cut vertically, it forms two identical plano-convex lenses. For a plano-convex lens, one surface is flat ( $R_{1} = \infty$ ) and the other has a radius of curvature $R_{2} = - R$ . The focal length $f_{o}^{'}$ is given by:

\frac{1}{f_{o}^{'}} = (μ - 1) (\frac{1}{\infty} - \frac{1}{- R}) = (μ - 1) \frac{1}{R}

Thus, $f_{o}^{'} = \frac{R}{μ - 1}$ . Comparing the two focal lengths, we find $f_{o}^{'} = 2 f_{o}$ .

Step 3: Relate magnification, object distance, and tube length to find the change in tube length○ Expand

The approximate magnification of a compound microscope is given by $M \approx \frac{L}{f_{o}} \frac{D}{f_{e}}$ , where $L$ is the tube length, $f_{o}$ is the objective focal length, $D$ is the near point, and $f_{e}$ is the eyepiece focal length. The problem states that the total magnification $M$ is to be retained, the object distance $u_{o}$ is unchanged (which implies the objective magnification $M_{o} = v_{o} / u_{o}$ is constant, and for approximation $u_{o} \approx f_{o}$ is maintained), and the eyepiece is unchanged (so $f_{e}$ is constant). Therefore, the ratio $\frac{L}{f_{o}}$ must remain constant.

\frac{L_{n e w}}{f_{o}^{'}} = \frac{L_{o l d}}{f_{o}}

Substituting $f_{o}^{'} = 2 f_{o}$ from Step 2, we get:

L_{n e w} = L_{o l d} (\frac{f_{o}^{'}}{f_{o}}) = L_{o l d} (\frac{2 f_{o}}{f_{o}}) = 2 L_{o l d}

The tube length has to be increased two times.

💡 Teacher's Secret Hint

Remember that for a compound microscope, the object is placed very close to the focal point of the objective lens, and the image formed by the objective is approximately at the tube length from the objective.

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Physics Question 42 – JEE-MAIN 2026

Clue 1 of 3: Lens Properties

🧠 Full Solution Path