A compound microscope has a magnification of 30 . The focal length of the eye-piece is 5 cm If the final image is formed at the least distance of distinct vision (25 cm) the magnification produced by the objective is

Question

A compound microscope has a magnification of 30 . The focal length of the eye-piece is 5 cm If the final image is formed at the least distance of distinct vision (25 cm) the magnification produced by the objective is (A)  10  (B)  7.5  (C)  5  (D)  15

Tardigrade Admin · Accepted Answer

m=30 fe=5 cm m=mo × me m=mo ×(1+(D/fe)) 30=m0(1+(25/5)) ⇒ mo=5

Q. A compound microscope has a magnification of $30$ . The focal length of the eye-piece is $5 c m$ If the final image is formed at the least distance of distinct vision $(25 c m)$ the magnification produced by the objective is

$10$

$7.5$

$5$

$15$

$m = 30 f_{e} = 5 c m$
$m = m_{o} \times m_{e}$
$m = m_{o} \times (1 + \frac{D}{f _{e}})$
$30 = m_{0} (1 + \frac{25}{5})$
$\Rightarrow m_{o} = 5$

Q. A compound microscope has a magnification of 30 . The focal length of the eye-piece is 5cm If the final image is formed at the least distance of distinct vision (25cm) the magnification produced by the objective is

10

7.5

5

15

m=30fe​=5cm m=mo​×me​ m=mo​×(1+fe​D​) 30=m0​(1+525​) ⇒mo​=5

Q. A compound microscope has a magnification of $30$ . The focal length of the eye-piece is $5 c m$ If the final image is formed at the least distance of distinct vision $(25 c m)$ the magnification produced by the objective is

$10$

$7.5$

$5$

$15$

$m = 30 f_{e} = 5 c m$
$m = m_{o} \times m_{e}$
$m = m_{o} \times (1 + \frac{D}{f _{e}})$
$30 = m_{0} (1 + \frac{25}{5})$
$\Rightarrow m_{o} = 5$