Question

In the given figure a plano-concave lens is placed on a paper on which a flower is drawn. How far

above its actual position does the flower appear to be ?

Answer 1

Answer: 10 cm above

Explanation:

Given data in the figure:

Object distance, u = - 20 cm

Refractive index, µ₁ = 3/2 & µ₂ = 1

Radius of curvature, R = + 20 cm

To find: the image distance above its actual position

Solution:

Let the image distance be denoted as “v” cm.

 

Using the general equation of refraction through a curved surface,

[µ₂/v] – [µ1/u] = [(µ2 - µ1)/R]

Substituting the given values, we get

[1/v] – [-(3/2)/20] = [(1 – 3/2)/20]

⇒ 1/v = - [3/40] – [1/40]

⇒ 1/v = [-3-1]/40

⇒ 1/v = -4/40

⇒ 1/v = - 1/10

⇒ v = - 10 cm

Thus, the flower appears to be 10 cm above its actual position.