Question 219: Ray Optics

Question

In a microscope the objective is having focal length $f_o=2 \mathrm{~cm}$ and eye-piece is having focal length $f_e=4 \mathrm{~cm}$. The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is $\_\_\_\_$ .

Accepted Answer

In microscope normal adjustment refers to the condition where the final image is formed at infinity.
The total magnifying power ( M ) of a compound microscope in normal adjustment is the product of the linear magnification of the objective ( $\mathrm{m}_{\mathrm{o}}$ ) and the angular magnification of the eyepiece ( $\mathrm{m}_{\mathrm{e}}$ ):
$$ \mathrm{M}=\mathrm{m}_{\mathrm{o}} 	imes \mathrm{m}_{\mathrm{e}} $$
For an image at infinity:
Objective magnification $\mathrm{m}_{\mathrm{o}}=\frac{\mathrm{L}}{\mathrm{f}_{\mathrm{O}}}$, where L is the tube length.
Eyepiece magnification $\mathrm{m}_{\mathrm{e}}=\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}$, where D is the least distance of distinct vision (standard value is 25 cm ).
Thus, the total magnification formula is:
$$ M=\frac{L}{f_o} 	imes \frac{D}{f_e} $$
The focal length of objective is $\mathrm{f}_{\mathrm{o}}=2 \mathrm{~cm}$
The focal length of eyepiece is $\mathrm{f}_{\mathrm{e}}=4 \mathrm{~cm}$
The tube length is $\mathrm{L}=32 \mathrm{~cm}$
Substituting the values,
$$ M=\frac{32}{2} 	imes \frac{25}{4} 100 $$
Therefore, the total magnification of given microscope is 100 . Hence, the correct answer is $\mathbf{1 0 0}$.

Ray Optics

In a microscope the objective is having focal length f_o=2 \mathrm{~cm} and eye-piece is having focal length f_e=4 \mathrm{~cm}. The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is \_\_\_\_ .

Integer / Numeric Answer Type