Question

A concave mirror produces three times larger image of an object placed at a distance of 15 cm in front of it. Calculate the image distance and the focal length of the concave mirror.​

Answer 1

Answer:

so plz convert into your problem

Explanation:

Given parameters

The concave mirror produces three times enlarged image, so

m = 3

u = -10 cm

m=−vu

3 = (-v)/(-10)

v = +30

The distance between the mirror and image = 30 cm

From mirror formula, we know

1v+1u=1f

130+1−10=1f

(-2)/30 = 1/f

-1/15 = 1/f

f = – 15 cm

The radius of curvature R = 2f

∴ R = 2 × (- 15)

R = – 30 cm

Answer 2

Given:

$\sf{3ho=hi}$

$\sf{u=-15cm}$

Solution:

$\sf{3ho=hi}$

$\sf{{\dfrac{hi}{ho}}=3}$

$\sf{{\dfrac{hi}{ho}}={\dfrac{-v}{u}}}$

$\sf{{\dfrac{-v}{u}}=3}$

$\sf{{\dfrac{-v}{-15}}=3}$

$\sf{v=45cm}$

From mirror formula:

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}+{\dfrac{1}{u}}}$

$\sf{{\dfrac{1}{f}}={\dfrac{1}{45}}-{\dfrac{1}{15}}}$

$\sf{{\dfrac{1}{f}}={\dfrac{1-3}{45}}}$

$\sf{{\dfrac{1}{f}}=-{\dfrac{2}{45}}}$

$\sf{f=-}{\dfrac{45}{2}}$

$\sf{f=22.5cm}$

Therefore, if the object will be at 15cm from mirror with magnification=3, the image will form at 45cm behind the mirror and the focal length will be 22.5cm.