Chapter 23, Problem 86GP

To determine

Part (a) To Determine:

The nature of lens and its focal length if the image is at 8.20 cm in front of the lens.

Answer to Problem 86GP

Solution:

The lens will be diverging or concave and its focal length will be $f=-10.5\text{cm}$ also the nature of image will be virtual, if the image is at 8.20 cm in front of the lens.

Explanation of Solution

Given Info:

Object distance $u=+37.5\text{cm}$

Image distance $v=-8.20\text{cm}----\text{since image is formed on object's side}$

Let the focal length of lens be ‘f’

Formula Used:

Apply focal length formula for lens

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

Here, all alphabets are in their usual meanings.

Calculation:

Since Object and image both are on the same side of the lens and $u>v$. Hence, the given lens must be diverging or concave lens.

Apply focal length formula for lens.

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{\text{-8}\text{.20}\text{cm}}+\frac{1}{\text{+37}\text{.5}\text{cm}}\\ or,f=-10.5cm\end{array}$

Since the given lend is diverging lens and its focal length is negative. Hence, the image will always be virtual.

To determine

Part (b) To Determine:

The nature of lens and its focal length if the image is at 44.5 cm in front of the lens.

Answer to Problem 86GP

Solution:

The lens will be converging or convex and its focal length will be $f=+238.4\text{cm}$ also the nature of image will be virtual, if the image is at 8.20 cm in front of the lens.

Explanation of Solution

Given Info:

Object distance $u=+37.5\text{cm}$

Image distance $v=-8.20\text{cm}----\text{since image is formed on object's side}$

Let the focal length of lens be ‘f’

Formula Used:

Apply focal length formula for lens

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

Here, all alphabets are in their usual meanings.

Calculation:

Since Object and image both are on the same side of the lens and $u<v$. Hence, the given lens must be converging or convex lens.

Apply focal length formula for lens.

$\begin{array}{l}\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{\text{1}}{\text{-44}\text{.5}\text{cm}}\text{+}\frac{\text{1}}{\text{+37}\text{.5}\text{cm}}\\ or,f=+238.4cm\end{array}$

Since the given lend is converging lens and its focal length is positive also $u<v$. Hence the image will again be virtual.