Question

An object 4 centimetre in size is placed at 25 cm in front of the concave mirror of focal length 15 cm at what distance from the mirror would a screen be placed in order to obtain a sharp image find the nature and the size of image

Answer 1

Given :

• Object distance (u) = - 25cm
• Object distance is negative because the distance is towards the left in concave mirror.
• Focal length (f) = - 15cm
• Size of the object (o) = 4 cm

To Find:-

• Image distance (v) =?
• Size of the image (i) =?

Formula used :

Mirror formula,

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

Solution :

$\frac{1}{v} + ( - \frac{1}{25} ) = - \frac{1}{15}$

$\frac{1}{v} = ( - \frac{1}{15} + \frac{1}{25} )$

$\frac{1}{v} = \frac{ - 25 + 15}{15 \times 25} = - \frac{ 2}{75}$

Taking the reciprocal we have,

v = - 75/2

v = (- 37.5)cm.

Now,

$As \: magnification \: (m) = \frac{i}{u} = - \frac{v}{u}$

Therefore,

$\frac{i}{4} = - (\frac{ - 37.5}{ - 25} )\\ i \: = - (\frac{37.5 \times 4}{25} )\\ i \: = ( - 6)cm$

Hence,

The image is formed at the screen at a distance of 37.5 cm and the size of the image is 6cm and the negative sign of the image shows that the image formed is inverted and enlarged of this concave mirror.