Question

An object is placed 25 cm in front of a concave lens of focal length 25 cm. Find the location and nature of image.

Answer 1

hyy mate ✌️

It is given that,

Object distance from the lens, u = -25 cm

Focal length of the concave lens, f = -25 cm

Let v is the image distance. It can be calculated using lens formula as :

v = -12.5 cm

So, the image is location in front of the concave lens at a distance of 12.5 cm.

Magnification,

m = 0.5

So, the formed image is virtual and erect...

hope it will help you ❣️

please mark as brainliest!!!

Answer 2

$\huge \bf \pink{Solution.}$

Given, object distance, u = –25 cm

Focal length, f = –25 cm, image distance, v = ?

From lens formula,

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

So,

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

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

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

$v = \frac{ - 25 \times ( - 25)}{ - 25 - 25} = - 12.5 \: cm$

Nature of image

1. Virtual.

2. Erect.

3. On the same side of lens.

4. Diminished.