Magnification of a compound microscope is 30 . The focal length of eye-piece is 5 cm and the image is formed at a distance of ‘distinct vision of 25 cm . Magnification of the objective lens is

Question

Magnification of a compound microscope is 30 . The focal length of eye-piece is 5 cm and the image is formed at a distance of ‘distinct vision of 25 cm . Magnification of the objective lens is (A) 6 (B) 5 (C) 7.5 (D) 10

Tardigrade Admin · Accepted Answer

m=m1⋅ m2=30 for eye-piece (1/v1)-(1/u)=(1/f) ⇒ (- 1/25)-(1/u)=(1/5) ∴ u=(- 25/6) ∴ m2=(- 25 × 6/- 25)=6 ∴ m1=5

Q. Magnification of a compound microscope is $30$ . The focal length of eye-piece is $5 c m$ and the image is formed at a distance of ‘distinct vision of $25 c m$ . Magnification of the objective lens is

$6$

$5$

$7.5$

$10$

$m = m_{1} \cdot m_{2} = 30$
for eye-piece $\frac{1}{v _{1}} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{- 1}{25} - \frac{1}{u} = \frac{1}{5}$
$∴ u = \frac{- 25}{6}$
$∴ m_{2} = \frac{- 25 \times 6}{- 25} = 6$
$∴ m_{1} = 5$

Q. Magnification of a compound microscope is 30 . The focal length of eye-piece is 5cm and the image is formed at a distance of ‘distinct vision of 25cm . Magnification of the objective lens is

6

5

7.5

10

m=m1​⋅m2​=30 for eye-piece v1​1​−u1​=f1​ ⇒25−1​−u1​=51​ ∴u=6−25​ ∴m2​=−25−25×6​=6 ∴m1​=5

Q. Magnification of a compound microscope is $30$ . The focal length of eye-piece is $5 c m$ and the image is formed at a distance of ‘distinct vision of $25 c m$ . Magnification of the objective lens is

$6$

$5$

$7.5$

$10$

$m = m_{1} \cdot m_{2} = 30$
for eye-piece $\frac{1}{v _{1}} - \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{- 1}{25} - \frac{1}{u} = \frac{1}{5}$
$∴ u = \frac{- 25}{6}$
$∴ m_{2} = \frac{- 25 \times 6}{- 25} = 6$
$∴ m_{1} = 5$