Chapter 23, Problem 93GP

To determine

To find the position of final image formed by the combination of lens and spherical mirror.

Answer to Problem 93GP

Solution:

The position of final image formed by the combination of lens and spherical mirror will be 6.0 cm from the main object towards the lens.

Explanation of Solution

Given Info:

Object distance $u=+30.0\text{}\text{cm}$

Power of lens $P=+5.0D\text{so,}\text{lense}\text{must}\text{be}\text{convex}$

Focal length of spherical mirror $f_{mirror}=25.0\text{cm}$

Distance of spherical mirror from the lens $d=75.0\text{cm}$

Let the focal length of lens be ‘f’

Formula Used:

Apply focal length formula for spherical mirror

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Here, all alphabets are in their usual meanings.

Calculation:

First, we shall calculate the focal length of lens as

$\begin{array}{l}f=\frac{1}{P}\\ f=\frac{1}{+5.0D}\\ f=+20.0\text{cm}\end{array}$

Now, we shall find the image distance so formed by convex lens as

Apply focal length formula for convex lens

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ or,\frac{1}{v}=\frac{1}{u}-\frac{1}{f}\\ or,\frac{1}{v}=\frac{\text{1}}{\text{+30}\text{.0}\text{cm}}\text{-}\frac{\text{1}}{\text{+20}\text{.0}\text{cm}}\\ or,v=-60.0\text{cm}\end{array}$

This tells that image formed by convex lens is formed behind the lens and is inverted and real.

Since this image will work as an object for the spherical mirror and mirror reflect light back through lens. Hence, spherical mirror must be concave mirror.

Hence the object distance for the mirror will be $u_{mirror}$ which can be calculated as

$\begin{array}{l}{u}_{mirror}=d-v\\ u_{mirror}=75.0\text{cm-60}\text{.0}\text{cm}\\ u_{mirror}=+15.\text{0}\text{cm}\end{array}$

Apply focal length formula for mirror,

$\begin{array}{l}\frac{1}{f_{mirror}}=\frac{1}{v_{mirror}}+\frac{1}{u_{mirror}}\\ or,\frac{1}{v_{mirror}}=\frac{1}{u_{mirror}}-\frac{1}{f_{mirror}}\\ or,\frac{1}{v_{mirror}}=\frac{\text{1}}{\text{+15}\text{.0}\text{cm}}\text{-}\frac{\text{1}}{\text{+25}\text{.0}\text{cm}}\\ or,v_{mirror}=\text{-37}\text{.5}\text{cm}\end{array}$

This image is formed at 37.5 cm behind the concave mirror which will works as an object for the convex lens because mirror reflects light through the lens.

Hence, we shall calculate again the object distance for the lens as

$\begin{array}{l}{u}_{lens}=d+v_{mirror}\\ u_{lens}=75.0\text{cm+37}\text{.5}\text{cm}\\ u_{lens}=+112.5\text{cm}\end{array}$

Let the final image distance for the convex lens be $v_{lens}$

Again, apply focal length formula for convex lens

$\begin{array}{l}\frac{1}{f}=\frac{1}{v_{lens}}+\frac{1}{u_{lens}}\\ or,\frac{1}{v_{lens}}=\frac{1}{u_{lens}}-\frac{1}{f}\\ or,\frac{1}{v_{lens}}=\frac{\text{1}}{\text{+112}\text{.5}\text{cm}}\text{-}\frac{\text{1}}{\text{+20}\text{.0}\text{cm}}\\ or,v_{lens}=-2\text{4}\text{.3}\text{cm}\\ or,v_{lens}\approx -24.\text{0}\text{cm}\end{array}$

This is the final image distance from the lens which will works as virtual image for the lens at it will be at $\left(v-v_{lens}\right)=3\text{0}\text{.0 }\text{cm-24}\text{.0 }\text{cm}=6.0cm$ distance from the main

object towards the lens.