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

Explanation:

Given:

The size of object = 4.0cm

Distance of the object = 25cm (-ve as it is in front of mirrors)

focal length of concave mirror = 15cm (-ve)

Distance of the image is given by-

v

1

+

u

1

=

f

1

v

1

+

−25

1

=

−15

1

v

1

=

−15

1

+

25

1

=

150

−10+6

150

−4

⇒v=−37.5cm

The screen should be at 37.5cm in front of the lens.

(ii)

The size of the image is given by-

h

0

h

i

=

u

−v

⇒h

i

=

u

−v×h

0

=

−25

−(−37.5)×4

h

i

=−

25

150

=−6cm

The image will be 6cm high and it will be inverted.

(iii)

Refer image 1

The image will be formed at a distance of 37.5cm from the mirror.

It will be an inverted image.

Answer 2

Explanation:

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 = (- 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.