Question

Question:-

In a compound microscope focal length of objectives lens is 1.2cm & focal length of eye piece is 3.0 cm when objects is kept at 1.25 cm infront of objective , final image is formed at ∞ . Magnifying power is ​

Answer 1

Given,

Fe = 3 cm

Fo = 1.2 cm

Uo= 1.25 cm

Ve= ∞

From,

$\frac{1}{uo} + \frac{1}{vo} = \frac{1}{fo}$

We get,

$\frac{1}{vo} = \frac{1}{1.2} - \frac{1}{1.25}$

$vo = 30cm$

Magnification of microscope m =

$\frac{vo}{uo} ( \frac{D}{fe} )$

Therefore, m =

$\frac{30}{1.25} ( \frac{25}{3}) \: \: \: \: \: = 200.$

Answer 2

Given :

• Focal length of objective lens (fo)= 1.2 cm

• focal length of eye piece (fe)= 3.0 cm

• objects is kept at 1.25 cm infront of objective (i.e u = 1.25 cm)

To find :

• Magnifying power of the compound microscope when final image is formed at infinity.

Formula used :

Lens Formula :-

$\sf \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Magnification :-

$\sf \: m = \dfrac{v_o}{u_o} \times \dfrac{D}{f_e}$

Solution :

First by using lens formula we will find Vo.

$\implies\sf \dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$

$\implies\sf \dfrac{1}{v_o} - \dfrac{1}{( - 1.25)} = \dfrac{1}{1.2}$

$\implies\sf \dfrac{1}{v_o} + \dfrac{1}{ 1.25} = \dfrac{1}{1.2}$

$\implies\sf \dfrac{1}{v_o} = \dfrac{1}{1.2} - \dfrac{1}{ 1.25}$

$\implies\sf \dfrac{1}{v_o} = \dfrac{10}{12} - \dfrac{100}{ 125}$

LCM of 12 and 125 = 12× 125 = 1500

$\implies\sf \dfrac{1}{v_o} = \dfrac{1250 - 1200}{1500}$

$\implies\sf \dfrac{1}{v_o} = \dfrac{50}{1500}$

$\implies\sf \dfrac{1}{v_o} = \dfrac{5 \cancel0}{150\cancel0}$

$\implies\sf \dfrac{1}{v_o} = \dfrac{ \cancel5 \: \: \: 1}{\cancel{150} \: \: \: \: 30} = \dfrac{1}{30}$

therefore , Vo = 30 cm

Now we will find magnification :-

$\sf \: m = \dfrac{30}{1.25} \times \dfrac{25}{3}$

(D= 25)

$\implies\sf \: m = \dfrac{30}{125} \times \dfrac{25}{3} \times 100$

$\sf \implies\: m = \dfrac{ \cancel{30} \: \: 10}{ \cancel{125} \: \: 5} \times \dfrac{\cancel{25} \: \: 1}{ \cancel3 \: \: 1} \times 100 = \dfrac{10}{5} \times 100$

$\implies \sf m = \dfrac{ \cancel{10} \: \: 2}{\cancel5 \: \: 1} \times 100 = 200$

Answer : 200