Question

a convex mirror of radius of curvature 40 cm can produce image 2 times as large as object. what would be the position of the object?​

Answer 1

A covex mirror of radius of curvature 40 cm produces an image which is 2 times as large as object.

• Radius of curvature (R) = 40 cm

Now,

=> R = 2f

=> 40 = 2f

=> f = 20 cm

Let height of image be h1 and height of object be h2.

According to question,

=> h1 = 2 h2

=> h1/h2 = 2 _____ (eq 1)

We know that..

=> h1/h2 = -v/u

=> 2 = -v/u [From (eq 1)]

=> 2u = - v

=> - v = 2u

According to mirror formula

=> 1/f = 1/v + 1/u

Substitute the known values in above formula

=> 1/20 = 1/(-2u) + 1/u

=> 1/20 = -1/2u + 1/u

=> 1/20 = (- 1 + 2)/2u

=> 1/20 = 1/2u

Cross multiply them

=> 2u = 20

=> u = 10 cm

(But we know that object always put on left side.. and left side it's value is negative)

$\huge{\bold{Correct\:Question\::}}$

A convex mirror of radius of curvature 40 cm can produce image 1/2 times as large as object. what would be the position of the object?

$\huge{\bold{Solution\::}}$

• Radius of curvature (R) = 40 cm

Now,

=> R = 2f

=> 40 = 2f

=> f = 20 cm

Let height of image be h1 and height of object be h2.

According to question,

=> h1 = 1/2 h2

=> h1/h2 = 1/2 _____ (eq 1)

We know that..

=> h1/h2 = -v/u

=> -v/u = 1/2

=> u = - 2v

From Mirror Formula

1/f = 1/v + 1/u

Substitute the known values in above formula

=> 1/20 = 1/v + 1/(-2v)

=> 1/20 = 1/v - 1/2v

=> 1/20 = (2 - 1)/2v

=> 1/20 = 1/2v

=> 2v = 20

=> v = 10 cm

Again by using mirror formula

=> 1/f = 1/v + 1/u

=> 1/20 = 1/10 + 1/u

=> 1/u = 1/20 - 1/10

=> 1/u = (1 - 2)/20

=> 1/u = -1/20

=> u = - 20 cm

∴ Position of the object is -20 cm from the mirror.

Answer 2

R = 40 cm

f = 20 cm

hi = 2 ho

hi/ho = 1/2

hi/ho = -v/u

-v/u = 1/2

u = - 2v

1/f = 1/v + 1/u

1/20 = 1/v + 1/(-2v)

1/20 = (2 - 1)/2v

1/20 = 1/2v

v = 10 cm

Again

1/f = 1/v + 1/u

1/20 = 1/10 + 1/u

1/u = 1/20 - 1/10

1/u = -1/20

u = - 20 cm