Question

A positive lens has a focal length of 10 cm. An object is located 25 cm from the lens.

a. Calculate how far the image is from the lens.

b. Determine whether the image is real or virtual.

c. Calculate the magnification of the image (state whether the image is erect or inverted).​

Answer 1

Answer:

a)We have,

$u = - 25 \: cm$

and,

$f = 10cm$

By using lens formula,we have,

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

So,

$\frac{1}{v } - \frac{1}{ - 25} = \frac{1}{10}$

$\frac{1}{v} = \frac{1}{10} - \frac{1}{25}$

$\frac{1}{v} = \frac{3}{50}$

$v = \frac{50}{3} = 16.66$

So,

$v = 16.66 \: cm$

Hence,

The image is situated at a distance 16.66 cm from the lens.

b)Since the focal length of the given lens is positive, by convention ,the lens is convex lens .

Here since,the object is placed at a distance beyond 2f from the convex lens, the image will be formed between f and 2f on the opposite side of the lens and the image formed will be real.

c)We know that,

$magnification \: of \: lens= \frac{v}{u}$

So,

$magnification \: of \: lens = \frac{ \frac{50}{3} }{ -25 }$

$= \frac{50}{3} \times ( \frac{ - 1}{25} ) = \frac{ - 2}{3} = - 0.66$

Since the magnification is negative, the image formed is real and inverted