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  • The eye-piece and objective of a microscope, having focal-lengths of 0.3 m and 0.4 m respectively, are separated by a distance of 0.2 m. Now the eye piece and the objective are to be interchanged s

The eye-piece and objective of a microscope, having focal-lengths of 0.3 m and 0.4 m respectively, are separated by a distance of 0.2 m. Now the eye piece and the objective are to be interchanged such that the angular magnification of the instrument remains same.The new separation between the lenses is:

Option: 1

0.4 \mathrm{~m}


Option: 2

0.6 \mathrm{~m}


Option: 3

0.8 \mathrm{~m}


Option: 4

0.9 \mathrm{~m}


Answers (1)

best_answer

Suppose the microscope forms the final image of the object at \infty or we are seeing the final image in the microscope with relaxed eye. Let L be the initial distance between objective and eyepiece of the microscope. Angular magnification produced by microscope in its normal adjustment, is given by

\mathrm{M=-\frac{v_{0}}{u_{0}} \times \frac{D}{f_{e}}}

In microscope, as the object is placed very near to focus of objective, hence
\mathrm{u}_{0}=\mathrm{f}_{0}$ and $\mathrm{v}_{0}=\mathrm{L}-\mathrm{u}_{\mathrm{e}}=\mathrm{L}-\mathrm{f}_{\mathrm{e}} \quad\left(\because \mathrm{u}_{\mathrm{e}}=\mathrm{f}_{\mathrm{e}}\right., for final image to be formed at \mathrm{\infty})
\mathrm{\therefore \quad M=\left[\left(L-f_{e}\right) / f_{0}\right] \times\left(D / f_{e}\right)} -----(1)

When the lenses are interchanged, let the new separation between the lenses be L’. Hence angular magnification, now becomes
\mathrm{M^{\prime}=-\frac{\left(L^{\prime}-f_{0}\right)}{f_{e}} \times \frac{D}{f_{0}}}

According to question, \quad M=M^{\prime}
\mathrm{-\frac{\left(L-f_{e}\right)}{f_{0}} \times \frac{D}{f_{e}}=-\frac{\left(L^{\prime}-f_{0}\right)}{f_{e}} \times \frac{D}{f_{0}} \quad \text { or } \quad L-f_{e}=L^{\prime}-f_{0}}
\mathrm{or\, L^{\prime}=L-f_{e}+f_{0}=0.2+0.3+0.4=0.9 m}

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