Question

An object 5cm in length is placed at a distance of 20cm in a front of a concave mirror of radius of
curvature 30cm. Find the position of the image its nature and size.

Answer 1

Explanation:

It will be found to be equal to 6 cm. Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm.

Answer 2

Given:-

• Radius of curvature (R) = 30 cm
• f = R/2 = 30/2 = 15 cm

• u = -20 cm

• h= 5 cm.

To Find:-

• the position of the image, its nature and size.

Solution:-

We Know that

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

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

$= > \frac{1}{v} = \frac{1}{15} - \frac{1}{ - 20}$

$= > v = 8.57cm$

∴ Image is virtual and erect and formed behind the mirror.

So,

$m = \frac{ - v}{u}$

or

$m = \frac{h_{i}}{ h_{o} }$

$= > \frac{h_{i}}{5} = \frac{8.57}{20} = 0.428$

$= > h_{i} = 0.428 \times 5 = 2.14cm$

Position of Image: Behind the mirror.

Nature of Image: Virtual and Erect.

Size of the Image: Diminished.