Q25E

Question

Repeat Exercise 34.24 for the case in which the end of the rod is ground to a concave hemispherical surface with a radius of 4.00 cm.

Step-by-Step Solution

Verified

Answer

The position of the image is 8.35cm to the left of the vertex.

The height of the image is 0.326mm.

The image is erect.

1Step 1: Image Distance

Object-Image relationship for spherical reflecting surfaces:

$\frac{n_{1}}{s} + \frac{n_{2}}{s} = \frac{n_{2} - n_{1}}{R}$

$n_{1}, n_{2} \Rightarrow$ The refraction index of both the surfaces, $s \Rightarrow$ the image distance from the vertex of the spherical surfaces, $s \Rightarrow$ the object distance from the vertex of the spherical surface, $R \Rightarrow$ The Radius of the spherical surface

$n_{1} = 1 (a i r) n_{2} = 1.6 (g l a s s)$

R is negative because it is in the opposite direction of refracted light

s = 0.24m

$R = - 0.04$

$\frac{n_{1}}{s} + \frac{n_{2}}{s} = \frac{n_{2} - n_{1}}{R} \frac{n_{2}}{s} = \frac{n_{2} - n_{1}}{R} - \frac{n_{1}}{s}$

$s' = \frac{n_{2}}{(\frac{n_{2} - n_{1}}{R} - \frac{n_{1}}{s})} = \frac{1.6}{(\frac{1.6 - 1}{- 0.04 m} - \frac{1}{0.24 m})} = - 0.0835 = - 8.35 c m p o s i t i o n : 8.35 c m$

2Step 2: Lateral Magnification

Lateral magnification for spherical refracting surfaces:

$m = - \frac{n_{1'_{s}}}{n_{1 s}} = \frac{y'}{y}$

m = The magnification, $n_{1}, n_{2} = >$ The refraction index of both the surfaces, $s^{'} = >$ the image distance from the vertex of the spherical surfaces, the object distance from the vertex of the spherical surface, $y^{'} = >$ The height of the image, y => The height of the object, m => is (+) when the image is erect and (-) when the image is inverted.