Question

A concave spherical surface of refractive index 3/2 is immersed in water of refractive index 4/ 3 if a point object lies in water at a distance of 10 cm from the pole of the reflective surface calculate the position of the image give the radius of curvature of spherical surface is 18cm.
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Answer 1

Answer:

Given :   R=−10 cm             n2=3/2               n1=4/3

Let the object is placed at a distance x in front of the spherical surface  i.e.  u=−x

Using:    vn2−un1=Rn2−n1

∴    v3/2−−x4/3=−103/2−4/3 

⟹  v=0.017x+1.33−1.5x

⟹   v<0  i.e. the image formed is always virtual in nature.

Answer 2

The position of the image is 10.5 cm.

Given, Refractive index of the spherical surface (μ₂) = 3/2

          Refractive index of water (μ₁) = 4/3

          Distance of the object from the pole (u) = 10 cm

          The radius of curvature of the spherical surface (R) = 18 cm

Suppose, the distance of the image = v

According to the Lens makers' formula,

μ₂/v - μ₁/v = (μ₂ - μ₁) / R

⇒ 3/2 v - 4/3 × 10 = (3/2 - 4/3) /18

⇒ 3/2 v - 2/15 = 1/18×6

⇒ v = 10.5

So, the distance of the image is 10.5 cm.