A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by 1 cm. To what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is (4/3) )

Question

A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by 1 cm. To what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is (4/3) ) (A)  1 cm (B)  (4/3) cm  (C)  3 cm (D)  4 cm

Tardigrade Admin · Accepted Answer

Water should be poured to a height h cm so that apparent shift in position of coin is 1 cm. As y=(1-(1/n)) h ⇒ h=(y/(1-(1)n))=(1/(1-(1)4 / 3))=(1/(1)4)=4 cm

Q. A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by $1 c m$ . To what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is $\frac{4}{3}$ )

$1$ cm

$\frac{4}{3} c m$

$3$ cm

$4$ cm

Water should be poured to a height $h c m$ so that apparent shift in position of coin is $1 c m$ . As $y = (1 - \frac{1}{n}) h$
$\Rightarrow h = \frac{y}{( 1 - \frac{1}{n} )} = \frac{1}{( 1 - \frac{1}{4/3} )} = \frac{1}{\frac{1}{4}} = 4 c m$

Q. A microscope is focused on a coin lying at the bottom of a beaker. The microscope is now raised up by 1cm. To what depth should the water be poured into the beaker so that coin is again in focus? (Refractive index of water is 34​ )

1 cm

34​cm

3 cm

4 cm