Question

38.
In a compound microscope focal length of objective
and eyepiece lens is 4 cm and 8cm respectively. If
length of tube is 30 cm then magnifying power for
normal adjustment will be:-
(1) 20.3 (2) 14.06 (3) 23.43 (4) 15.3​

Answer 1

The magnification power is 23.43  

Explanation:

Given as :

For Compound microscope

The focal length of objective lens = $F_o$ = 4 cm

The focal length of eye piece lens = $F_e$ = 8 cm

The length of the tube = L = 30 cm

The least distance for clear vision = D = 25 cm

Let The magnification power = M

According to question

∵   Magnification power = ( - $\dfrac{L}{F_o}$  ) ( $\dfrac{D}{F_e}$ )

Or,                               M =   ( - $\dfrac{30 cm}{4 cm}$  ) ( $\dfrac{25 cm}{8 cm}$ )

Or,                               M = $\dfrac{750}{-32}$

∴                                   M = - 23.43

So, The magnification power = M = 23.43

Hence, The magnification power is 23.43  Answer

Answer 2

Explanation:

The question has error. It's of 2016 but the question seems to be wrong