Question

An object 4 cm in size is placed at 25 cm In front of a concave mirror of focal length 15 cm . At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image?

Answer 1

Answer:

ho=4cm

f= -15cm

u= -25cm

1/f=1/v+1/u

1/f-1/u=1/v

1/-15-1/-25=1/v

-1/15+1/25=1/v

-5/75+3/75=1/v

-2/75=1/v

v=-75/2cm

v=-37.5cm

this shows that the image is real ,inverted and enlarged

screen should be placed left side of the mirror at a distance of 37.5cm from the mirror

pls mark as brainliest.

Answer 2

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.