Question

An object 4cm in size is placed at a distance of 25cm in front of a convex mirror of radius of curvature 40 cm. Find the position the size the nature of the object.

Answer 1

Answer:

The image will form 11.11cm behind the mirror and is diminished

the size of the image is 1.77cm

Explanation:

Let us consider

• object distance as u

• image distance as v

• radius of curvature as r

• focal distance as f

• size of the image be i

• size of the object be o

Given data according to sign convention :

u = -25cm

v = ?

r = 40cm

f = ?

but we have ,

f = 1/2×r

→f = r/2

→f = 40/2

→f = 20cm

Now we have from mirror formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \implies \frac{1}{v} + \frac{1}{ - 25} = \frac{1}{20} \\ \implies \frac{1}{v} = \frac{1}{20} + \frac{1}{25} \\ \implies \frac{1}{v} = \frac{5 + 4}{100} \\ \implies \frac{1}{v} = \frac{9}{100} \\ \implies v = \frac{100}{9} \\ \implies v = 11.11cm$

★Therefore, the image distance is +11.11cm

here , the '+' sign signify that the image is formed 11.11cm behind the mirror

and magnification , m = i/o = - v/u

$\frac{i}{o} = - \frac{v}{u} \\ \implies \frac{i}{4cm} = \frac{ - 11.11cm}{ - 25cm} \\ \implies i = \frac{ 11.11 \times 4cm}{25} \\ \implies i = 1.776cm$

Size of the image formed is 1.77 cm

the image will be virtual and erect

Answer 2

the image will be virtual and erected as the image distance is positive

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