Question

An object 4.00cm in size ,is placed at 25.0 cm in front of a concave mirror of focal lenth 15.0cm . 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:

Explanation:

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The size of the image is 6cm

The nature of the image is real and inverted and enlarged and distance is -37.5 cm.

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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.