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Magnification of an astronomical telescope is 4 for normal vision and its tube length for normal vision is 1m. What will be the tube length if the final image is formed at least distance for distinct vision (25 cm)?

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75 cm

88 cm

85 cm

91 cm

answer is D.

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Detailed Solution

For normal vision, $\frac{f_{o}}{f_{e}} = 4 and f_{0} + f_{e} = 100 c m$

Solving, $f_{e} = 20 c m and f_{0} = 80 c m$

When the image is formed at least distance, for distinct vision,

$L = f_{0} + \frac{D f_{e}}{D + f_{e}} = 80 + (\frac{25 \times 20}{25 + 20}) C m ≃ 91 c m$

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