A glass sphere having refractive index 3 / 2 is having a small irregularity at its centre. It is placed in a liquid of refractive index 4 3 such that the surface of the liquid is at a distance r above the sphere, where r = 20 cm is radius of the sphere. If the irregularity is viewed from above then what is it's distance (in cm ) from the centre where eye will observe the irregularity?

Question

A glass sphere having refractive index 3 / 2 is having a small irregularity at its centre. It is placed in a liquid of refractive index 4 3 such that the surface of the liquid is at a distance r above the sphere, where r = 20 cm is radius of the sphere. If the irregularity is viewed from above then what is it's distance (in cm ) from the centre where eye will observe the irregularity?,

MARKS App · Accepted Answer

Consideration refraction at glass-water interface. μ 2 v - μ 1 u = μ 2 - μ 1 R ⇒ 4 3 v - 3 - 2 r = 4 / 3 - 3 / 2 r ⇒ 4 3 v + 3 2 r = - 1 6 r ∴ v = r 5 Now refraction at water air surface u = - r + 4 5 r = - 9 r 5 μ 2 v - μ 1 u = μ 2 - μ 1 ∞ ⇒ 1 v + 4 × 5 3 × 9 r = 1 - 4 / 3 ∞ 1 v = - 20 27 r v = - 27 r 20 So height above center = 2 r - 27 r 20 = 40 r - 27 r 20 = 13 20 r = 13 cm

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