Question

The left end of a long glass rod 8.20 cm in diameter, with an index of refraction 1.58, is ground and polished to a convex hemispherical surface with a radius of 4.10 cm . An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cmto the left of the vertex of the convex surface.

1)

Part C

Is the image erect or inverted?

Answer 1

Given

Refractive index of air n1 = 1

Refractive index of the Glass rod n2 = 1.58

Length of the Rod d = 7 cm

Radius of the hemispherical surface r = + 4.10 cm

Distance of the Object u = + 24.0 cm

Solution

(n₁/u) +(n₂/v) = n₂-n₁/r

(1/24) + (1.58/v) = (1.58-1)/4.1

1.58/v = (0.58/4.1) – (1/24)

1.58/v= 0.1

V = 15.8

Yes the value of image distance v is positive hence both the object and image appear on the same side which means a real invented image has been formed