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Q.

A microscope is used to resolve two self luminous objects separated by a distance of 5 × 10^–5cm and placed in air. What should be the minimum magnifying power of the microscope in order to see the resolved images of the objects? The resolving power of the eye is 1.5 and the least distance of distinct vision is 25 cm.
$(Hint : Δx = DΔθ and M = \frac{Δx}{d})$

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a

100

b

400

c

250

d

220

answer is C.

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Detailed Solution

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$\begin{array}{l} Δx = DΔθ and M = \frac{Δx}{d} \\ M = \frac{aperture of the objective}{aperture of the eye} = \frac{a_{0}}{a_{e}} \\ Δθ = \frac{1 .22 λ}{a_{0}}, a_{0} = \frac{1 .22 λ}{Δθ} = \frac{1 .22 λD}{Δx} \\ d = \frac{1 .22 λ}{a_{e}}, a_{e} = \frac{1 .22 λ}{d}; M = \frac{Δd}{x} = \frac{25 \times 1 .5}{5 \times 10^{- 5}} \\ M = \frac{25 \times 1 .5 \times π}{60 \times 5 \times 10^{- 5} \times 180} = 220 \end{array}$

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A microscope is used to resolve two self luminous objects separated by a distance of 5 × 10–5 cm and placed in air. What should be the minimum magnifying power of the microscope in order to see the resolved images of the objects? The resolving power of the eye is 1.5 and the least distance of distinct vision is 25 cm. (Hint: Δx=DΔθ and M=Δxd