The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.70 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
B) Find the height of the image formed by paraxial rays incident on the convex surface.
C) Is the image erect or inverted?

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.