Question

The left end of a long glass rod 9.00 cmcm in diameter, with an index of refraction 1.55, is ground and polished to a convex hemispherical surface with a radius of 4.50 cmcm. An object in the form of an arrow 1.52 mmmm tall, at right angles to the axis of the rod, is located on the axis 25.0 cmcm 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. (answer is s1 in units cm)
B) Find the height of the image formed by paraxial rays incident on the convex surface. (answer is y1 in units mm)
C) Is the image erect or inverted?
I have tried solving this problem multiple ways and even referred to previous answers, but nothing seems to work. I will rate lifesaver for anyone who shows a correct explanation of how to solve and approach this problem. I greatly appreciate any help you can provide!

Answer 1

A) s' = 18.85 cm

B) Height is 0.739 mm

C) the image is inverted.

Step-by-step explanation:

We are given;

Diameter; D = 9 cm

Radius; R = d/2 = 9/2 = 4.5 cm = 0.045 m

Refractive Index of glass; n₂= 1.55

Height of object; y = 1.52 mm = 0.00152 m

Object distance s = 25 cm = 0.25 m

Now, Refractive index of air is 1 from online values.

Thus; n_1 = 1

A) To find the position of the image of the arrow formed by paraxial rays incident on the convex surface which is denoted by s', we will use the formula;

((n_1)/s) + n₂/s' = (n₂ - n_1)/R

Plugging in the relevant values, we have;

(1/0.25) + 1.55/s' = (1.55 - 1)/0.045

(1/0.25) + 1.55/s' = 12.222

1.55/s' = 12.222 - (1/0.25)

1.55/s' = 12.222 - 4

1.55/s' = 8.222

s' = 1.55/8.222

s' = 0.1885 m

s' = 18.85 cm

B) We will use the magnification formula to calculate the height of the image formed by paraxial rays incident on the convex surface denoted by y';

m = y'/y = -n_1•s'/n_2•s

Thus;

y' = -(y × n_1 × s')/(n_2•s)

y' = -(0.1885 × 1 × 0.00152)/(1.55 × 0.25)

y' = -0.739 mm

Height is 0.739 mm inverted.

C) Since the height of y' is negative, then it means the image is inverted.