Chapter 18, Problem 77A

a.

To determine

Distance of copy paper from lens.

Answer to Problem 77A

  $v=66.67\text{cm}$

Explanation of Solution

Given:

Focal length of convex lens, $f=25\text{cm}$

Distance of letter from lens, $u=-40\text{cm}$

Formula used:

The lens formula is given as,

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

Where, $f$ is focal length of lens, $u$ is object distance and $v$ is image distance.

Calculation:

On substituting the given values,

  $\begin{array}{l}\frac{1}{25}=\frac{1}{v}-\frac{1}{-40}\\ \Rightarrow \frac{1}{v}=\frac{1}{25}-\frac{1}{40}=\frac{40-25}{25\times 40}\\ \Rightarrow v=\frac{1000}{15}=66.67\text{cm}\end{array}$

Conclusion:

Therefore, the distance of the copy paper from lens is $v=66.67\text{cm}$ .

b.

To determine

Height of the copy.

Answer to Problem 77A

  $h_i=-1.67h_o$

Explanation of Solution

Given:

Focal length of convex lens, $f=25\text{cm}$

Distance of letter from lens, $u=-40\text{cm}$

Formula used:

From the lens formula,

  $\frac{h_i}{h_o}=\frac{v}{u}$

Here, $h_i$ is height of image, $h_o$ is height of object, $v$ is image distance and $u$ is object distance.

Calculation:

As, from previous subpart $v=66.67\text{cm}$

On substituting the value,

  $\begin{array}{l}\frac{h_i}{h_o}=\frac{66.67}{-40}=-1.67\\ \Rightarrow h_i=-1.67h_o\end{array}$

So the copy is 1.67 times of the original dimension and inverted.

Conclusion:

Therefore, the copy is 1.67 times of the original dimension and inverted.