Question

an object 4cm size is placed at 25cm in front of a cancave mirror of focal length 15cm at what distance from the mirror whould a screen be placed in order to obtain a sharp image find the nature and size of image​

Answer 1

Since the Distance is Negative, thus the Image is Real. Thus, Screen is to be placed at a Distance of 37.5 cm behind the Mirror.

Answer 2

Solution:

focal length=-15cm

object distance= -25cm

object height= +4cm

image distance=?

image height=?

Substitute the above values in the equation

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

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

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

$\frac{1}{v} = \frac{ - 2}{75}$

$= > v = - 37.5cm$

So the screen should be placed at 37.5 cm from the pole of the mirror.The image is real.

$magnification \: m = \frac{hi}{ho } = \frac{ - v}{u}$

By substituting the above values

$\frac{h}{4} = \frac{ - (37.5)}{( - 25)}$

$h_{i} = \frac{ - (37.5 \times 4)}{(25)}$

$h_{i} = - 6 \: cm$

So, the image is inverted and enlarged.