Question

An object 0.4 m high is placed at a distance of
0.8 m from a concave mirror of radius of curvature
0.4 m. Find the position, nature and size of the image
formed.​

Answer 1

Explanation:

r = 2f

f = r/2 = 0.4/2 = 0.2 m

1/v +1/u = 1/f

1/v + 1/0.8 = 1/0.2

1/v = 10/2 - 10/8 = 5-(5/4) = 15/4

v = 4/15 m

Now , M = -I/0.4 = (4/15)/0.8

Or, -I = -0.4*4*(0.8)/15

= -(1.6)*(0.8)/15

= -0.085 m

So, the image will be formed before concave mirror at a distance of 4/15 m & the size will be 0.085 m high , real & inverted .

Answer 2

Given:

An object 0.04 m high is placed at a distance of 0.8 m from a concave mirror of radius of curvature 0.4 m.

To find:

Formula used

Position of image

Image characteristics

Calculation:

Applying Mirror Formula:

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

$\sf \implies \dfrac{2}{r} = \dfrac{1}{v} + \dfrac{1}{u}$

$\sf \implies \dfrac{2}{( - 40)} = \dfrac{1}{v} + \dfrac{1}{ (- 80)}$

$\sf \implies - \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ 80}$

$\sf \implies \dfrac{1}{v} = \dfrac{1}{ 80} - \dfrac{1}{20}$

$\sf \implies \dfrac{1}{v} = \dfrac{1 - 4}{ 80}$

$\sf \implies \dfrac{1}{v} = \dfrac{ - 3}{ 80}$

$\sf \implies \: v = - 26.67 \: cm$

So, image distance is -26.67 cm.

$\rm \: mag. = - \dfrac{v}{u}$

$\rm \implies \: mag. = - \dfrac{ - 26.67}{ - 80}$

$\rm \implies \: mag. = - 0.33$

So, image is real , inverted and diminished.

Also, both A) and B) are correct.