SATHEE: Optics Question 131

Optics Question 131

Question: In a compound microscope, the focal length of the objective and the eye lens are 2.5 cm and 5 cm respectively. An object is placed at 3.75 cm before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be (i.e. length of the microscopic tube)

Options:

A) 11.67 cm

B) 12.67 cm

C) 13.00 cm

D) 12.00 cm

Show Answer

Answer:

Correct Answer: A

Solution:

$L_{D} = v_{o} + u_{e}$ and for objective lens $\frac{1}{f_{o}} = \frac{1}{v_{o}} - \frac{1}{u_{o}}$

Putting the values with proper sign convention. $\frac{1}{+ 2.5} = \frac{1}{v_{o}} - \frac{1}{(- 3.75)}$

$\Rightarrow v_{o} = 7.5 c m$ For eye lens $\frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}$

$\Rightarrow \frac{1}{+ 5} = \frac{1}{(- 25)} - \frac{1}{u_{e}}$

$\Rightarrow u_{e} = - 4.16 c m$

$\Rightarrow | u_{e} | = 4.16 c m$ Hence $L_{D} = 7.5 + 4.16 = 11.67 c m$

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Optics Question 131

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