Question

An object is placed at 25cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror a screen be placed in order to obtain a sharp image

Answer 1

1/f=1/u+1/vu=-25cm1/-15=1/-25+1/(+v)===>1/v=-1/15+1/251/v=-25+15/15*251/v=-2/75v=-75/2m=-(v/u)=-(75/2/-25/1)=-3/2=-1.5m=-1.5so Image is magnified and real ,inverted
OR
ho = 4 cm

u = - 25 cm

f = - 15 cm

1/u + 1/v = 1/f

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

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

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

1/v = -2/75

v = -75/2 = - 37.5 cm

hi / ho = -v/u

hi / 4 = 75/2 * (-25)

hi = -3/2 * 4

hi = -3 * 2 = -6 cm

Therefore the image is real inverted . It is formed 6cm away from the mirror in front of the mirror.

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Answer 2

Answer:

$\displaystyle \red{{ \text{h}_i=6 \ cm}$

Explanation:

Given :

Object distance u = 25 cm

Focal length f = 15 cm

We have mirror formula :

1 / f = 1 / v + 1 / u

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

1 / v = 1 / 25 - 1 / 15

5 / v = 1 / 5 - 1 / 3

5 / v = ( 3 - 5 ) / 15

5 / v = - 2 / 15

v = - 75 / 2 = - 37.5 cm

We know :

h_i / h_o = - v / u

Where i and o represent image and object

h_i = - ( - 37.5)  × 4 / 25

h_i = 6 cm

Nature of the image are as :

Real , inverted and magnified.