Question

An object is placed in air at a distance of 20cm from a convex surface of a medium of refractive index 1.6. If a virtual image is formed at a distance of 45cm from the surface, find the radius of curvature of the surface

Answer 1

Solution :-

Refractractive index of air medium n₁ = 1

Refractive index of medium of a convex surface n₂ = 1.6

Object distance u = - 20 cm

Image distance v = 45 cm

Let 'R' be the radius of curvature of the surface

Using Curved surface formula

n₂/v - n₁/u = (n₂ - n₁)/R

Substituting the values

$\implies \sf \dfrac{1.6}{45} - \dfrac{1}{ - 20} = \dfrac{1.6 - 1}{R}$

$\implies \sf \dfrac{16}{450} + \dfrac{1}{20} = \dfrac{0.6 }{R}$

$\implies \sf \dfrac{32 + 45}{900} = \dfrac{0.6 }{R}$

$\implies \sf \dfrac{77}{900} = \dfrac{0.6 }{R}$

$\implies \sf R= \dfrac{0.6 }{77} \times 900$

$\implies \sf R= \dfrac{540}{77}$

$\implies \sf R= \dfrac{540}{77}$

$\implies \sf R= 7 \dfrac{1}{77}$

Therefore the radius of curvature of the surface is 7 1/77 cm or 7.0129 cm.

Answer 2

Answer:

$\large\boxed{\sf{7.01\;\;cm}}$

Explanation:

We have been geiven that:

Object distance, u = -20 cm

°.° Surface is convex .

.°.Image distance, v = +45 cm

Refractive infex of air, $\bold{n_{1}=1}$

Refractive index of medium, $\bold{n_{2}=1.6}$

Let, the Radius of curvature = r

Now, We have curved spherical surface formula:

$\large \boxed { \red{\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{r} }}$

Putting the respective values, we get

$= > \frac{1.6}{45} - \frac{1}{ - 20} = \frac{1.6 - 1}{r} \\ \\ = > \frac{1 .6}{45} + \frac{1}{20} = \frac{0.6}{r} \\ \\ = > \frac{(1.6 \times 4) +( 1 \times 9)}{180} = \frac{0.6}{r} \\ \\ = > \frac{6.4 + 9}{180} = \frac{0.6}{r} \\ \\ = > r = \frac{0.6 \times 180}{15.4} \\ \\ = > r = \frac{108}{15.4} \\ \\ = > r = 7.01$

Hence, radius of curvature = 7.01 cm