Question

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end:
(a) infinitely far,
(b) 12.0 cm;
(c) 2.00 cm.

Answer 1

Complete Question

The left end of a long glass rod 6.00 cm in diameter has a convex hemispherical surface 3.00 cm in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances  1.00 cm  to the left of the vertex of the curved end:

(a) infinitely far,

(b) 12.0 cm;

(c) 2.00 cm.

Answer:

The  image distance is  $v = 2 \ cm$

Explanation:

From the question we are told  that

   The  diameter is  $d = 6.0 \ cm = 0.06 \ m$

   The radius of the convex hemisphere is  $r = 3.0 \ cm = 0.03 \ m$

    The  refractive index is $n = 1.60$

     The object distance is  $u = 2.00 \ cm = 0.02 \ m$

      The refractive index of air is  $n_a = 1$

Generally lens maker's law is mathematically represented as

        $\frac{n_a}{u} - \frac{n}{v} = \frac{(n - n_a )}{r}$

=>     $\frac{1}{1} - \frac{1.60}{v} = \frac{(1.60 - 1 )}{3}$

=>   $v = 2 \ cm$

Answer 2

The position of the image if an object is placed in air on the axis of the rod at the given distances to the left of the vertex are;

A) Infinitely far; s' = 8 cm

A) Infinitely far; s' = 8 cmB) At 12 cm; s' = 13.71 cm

A) Infinitely far; s' = 8 cmB) At 12 cm; s' = 13.71 cmC) At 2 cm; s' = -5.33 cm

We are given;

Refractive index of glass; η_g = 1.6

Radius of hemispherical surface; r = 3 cm

Diameter of glass rod; d = 6 cm

A) The formula for equation for refraction at a single surface to relate the image and object distance is;

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

Where;

n1 = n_a which is Refractive index of air = 1

n2 = n_g = 1.6

Thus at s = Infinity, we have;

(1/∞) + (1.6/s') = (1.6 - 1)/3

When a number divides infinity it equals zero.

Thus;

0 + (1.6/s') = 0.6/3

s' = 1.6/0.2

s' = 8 cm

B) At s = 12 cm, we have;

(1/12) + (1.6/s') = (1.6 - 1)/3

0.0833 + (1.6/s') = 0.2

(1.6/s') = 0.2 - 0.0833

s' = 1.6/0.1167

s' = 13.71 cm

C) At s = 2 cm, we have;

(1/2) + (1.6/s') = (1.6 - 1)/3

0.5 + (1.6/s') = 0.2

(1.6/s') = 0.2 - 0.5

s' = -1.6/0.3

s' = -5.33 cm

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