Question

What curvature must be given to the bounding surface of a refracting medium (u = 1.5) for the
virtual image of an object in the adjacent medium (u = 1) at 10 cm, to be formed at a distance of
40 cm?
(1) 15 cm
(2) 12 cm
(3) 10 cm
4) 8 cm ​

Answer 1

Answer :-

→ Radius of curvature is 8 cm .

To Find :-

→ Radius of curvature .

Explanation :-

Given that ,

$\star \: \sf{\mu_{1} = 1 \: (for \: adjacent \: medium)} \\ \star \: \sf{ \mu_{2} = 1.5 \: \: , \: v = - 40 \: cm \: ( \because \: virtual \: image)} \\ \star \: \sf{u = - 10 \: cm}$

R is radius of curvature .

We know that ,

$\to \sf{ \frac{- \mu_{1} }{u} + \frac{ \mu_{2} }{v} = \frac{ \mu_{1} - \mu_{2} }{R} }$

substitute the given values ,

$\leadsto \sf{ \frac{ - 1}{ - 10} + \frac{1.5}{ - 40} = \frac{1 - 1.5}{R} } \\ \\ \leadsto \sf{ \frac{1}{10} - \frac{1.5}{40} = \frac{ - 0.5}{R} } \\ \\ \leadsto \sf{ \frac{4 - 1.5}{40} = \frac{ - 0.5}{R} } \\ \\ \leadsto \sf{ \frac{2.5}{40} = \frac{ - 0.5}{R} } \\ \\ \leadsto \sf{ \frac{25}{40} = \frac{ - 5}{R} } \\ \\ \leadsto \sf{ R = - \frac{40 \times 5}{25} } \\ \\ \leadsto \boxed{ \sf{R = - 8 \: cm \: }}$

Negative sign is for direction . distance is not negative .

Radius of curvature is 8 cm.

Answer 2

☯ GiveN :

$\sf{\mu_1 = 1}$ (It is for adjacent medium)

$\sf{\mu_2 = 1.5}$ (It is for refracting medium)

v = -40 cm (Because image is virtual)

u = -10 cm

$\rule{200}{1}$

☯ To FinD :

We have to find the radius of curvature (r).

$\rule{200}{1}$

☯ SolutioN :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{ \dfrac{-\mu_1}{u} + \dfrac{\mu_2}{v} = \dfrac{ \mu_1 - \mu_2}{r} }}}}}$

★ Putting Values ★

$\sf{\dashrightarrow \frac{\cancel{-}1}{\cancel{-}10} + \frac{1.5}{-40} = \frac{1 - 1.5}{r}} \\ \\ \sf{\dashrightarrow \frac{1}{10} - \frac{1.5}{40} = \frac{-0.5}{r}} \\ \\ \sf{\dashrightarrow \frac{4 - 1.5}{40} = \frac{-0.5}{r}} \\ \\ \sf{\dashrightarrow \frac{2.5}{40 \times (-0.5)} = \frac{1}{r}} \\ \\ \sf{\dashrightarrow \frac{1}{r} = \frac{2.5}{-20}} \\ \\ \sf{\dashrightarrow r = \frac{-20}{2.5}} \\ \\ \sf{\dashrightarrow r = -8} \\ \\ \Large{\implies{\boxed{\boxed{\sf{r = -8 \: cm}}}}}$