Question

An object is placed a distance of twice the focal length away from a diverging lens. What is the magnification of the image?

Answer 1

The magnification of the image is -⅓

$\texttt{ }$

Further explanation

We will solve this problem using following formula:

$\boxed {\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}}$

$\boxed {M = s_i \div s_o}$

where:

so = distance of object from lens

si = distance of image from lens

f = focal length of lens

M = magnification of lens

$\texttt{ }$

Given:

focal length of diverging lens = -f

distance of object from lens = so = 2f

Asked:

magnification of the image = M = ?

Solution:

Firstly , we will find the distance of image from lens as follows :

$\frac{1}{s_o} + \frac{1}{s_i} = \frac{1}{f}$

$\frac{1}{2f} + \frac{1}{s_i} = \frac{1}{-f}$

$\frac{1}{s_i} = - \frac{1}{f} - \frac{1}{2f}$

$\frac{1}{s_i} = - \frac{3}{2f}$

$\boxed {s_i = - \frac{2}{3} f}$

$\texttt{ }$

Next , we could calculate the magnification of the lens as follows :

$M = s_i \div s_o$

$M = - (\frac{2}{3} f) \div ( 2f )$

$\boxed {M = - \frac{1}{3}}$

$\texttt{ }$

Conclusion :

The magnification of the image is -⅓

$\texttt{ }$

Learn more

• Compound Microscope : brainly.com/question/7512369
• Reflecting Telescope : brainly.com/question/12583524
• Focal Length : brainly.com/question/8679241
• Mirror an Lenses : brainly.com/question/3067085

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Light - Diverging Lens

Answer 2

Answer:

+1/3

Explanation:

The lens equation states that:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

For a diverging lens, the focal length is negative: $f=-f$

and we also know that the object is placed a distance of twice the focal length, so $p=2f$

So we can find q from the equation above

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-f}-\frac{1}{2f}=-\frac{3}{2f}\\q=-\frac{2}{3}f$

And the magnification of the image is given by

$M=-\frac{q}{p}=-\frac{-\frac{2}{3}f}{2f}=\frac{1}{3}$