Ray Optics

JEE Main level, single correct, medium

273 of 295

Question 273 of 295

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

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Answer: (A) increased two times

Focal length of the objective after cutting:
For a symmetric biconvex lens with both radii of magnitude RR:
1fo=(μ−1)(1R−1−R)=2(μ−1)R\dfrac{1}{f_o} = (\mu - 1)\left(\dfrac{1}{R} - \dfrac{1}{-R}\right) = \dfrac{2(\mu - 1)}{R}

When the lens is cut vertically (perpendicular to the principal axis), each half becomes a plano-convex lens with one flat surface:
1fo′=(μ−1)(1R−0)=μ−1R\dfrac{1}{f_o'} = (\mu - 1)\left(\dfrac{1}{R} - 0\right) = \dfrac{\mu - 1}{R}

Comparing the two: fo′=2fof_o' = 2f_o, so the focal length of the objective doubles.

Magnification of a compound microscope:
M≈Lfo⋅DfeM \approx \dfrac{L}{f_o} \cdot \dfrac{D}{f_e}

where LL is the tube length, fof_o is the objective focal length, fef_e is the eyepiece focal length and DD is the least distance of distinct vision.

Since the eyepiece and DD are unchanged, equating the initial and final magnifications:
Linitialfo⋅Dfe=Lfinal2fo⋅Dfe\dfrac{L_{\text{initial}}}{f_o} \cdot \dfrac{D}{f_e} = \dfrac{L_{\text{final}}}{2f_o} \cdot \dfrac{D}{f_e}

Linitialfo=Lfinal2fo\dfrac{L_{\text{initial}}}{f_o} = \dfrac{L_{\text{final}}}{2f_o}

Lfinal=2 LinitialL_{\text{final}} = 2\,L_{\text{initial}}

Hence, to maintain the same magnification, the tube length must be doubled.

The correct option is (1)(1) increased two times.

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