Question

A plastic hemisphere has a radius of curvature 8 cm and an index of refraction 1.6. On the axis halway between the plane surface and spherical one is a small object

Answer 1

Answer:

Explanation:

Given:

Index of refraction of plastic hemisphere, μ₁ = 1.6

Index of refraction of air, μ₂ = 1

Radius of curvature of plastic hemisphere, R = =-8 cm

Distance of small object, u = -4 cm

=> When viewed from plane surface:

μ₂/v -μ₁/u = μ₂-μ₁/R

1/v - 1.6/-4 = 1 - 1.6 / ∞ = 0

1/v + 1.6/4 = 0

1/v = -1.6 / 4

v = -4 / 1.6

v = - 2.5 cm

=> When viewed from curved surface:

μ₂/v -μ₁/u = μ₂-μ₁/R

1/v - 1.6/-4 = 1 - 1.6 / -8

1/v + 1.6/4 = 0.6/ 8

1/v = 0.6 / 8 - 1.6 / 4

1/v = 0.6 - 3.2 / 8

1/v = - 2.6 / 8

1/v = - 0.325

v = 1 / -0.325

v = - 3.076 cm

=> Distance between images :

= (8 - 2.5) - 3.076 cm

= 5.5 - 3.076 cm

= 2.43 cm

≈ 2.5 cm

Thus, the distance between the two images when viewed along the axis from the two sides of the hemisphere is approximately 2.5 cm.