Question

can anyone help with this one. I need right answer with full solution​

Answer 1

Answer:

given radius of curvature is = 20 cm

so focal length = 20/2 = 10cm

also ...

I= 1/2O

SO linear magnification = I / O = v / u = ( 1/2) O /O

= 1/2

SO

v / u = 1/ 2 ..

therefore u = 2v

so putting it in mirror formula

1/ u + 1 / v = 1/ f

1 / 2 v + 1/v = 1/ 10

3/2v = 1/ 10

2v / 3 = 10

v = 15 cm ( positive)

and u = 2 v = 2× 15 = 30 cm ( negative)

Answer 2

Explanation:

f = R/2 = +10cm ( in convex lens and mirrors the focal length is +ve)

h'/h = m = +1/2 ( image is virtual and erect in convex mirror)

h' is image height and h is object height

m= -v/u (v is image distance, u is object distance)

1/2 = -v/u -> u = -2v

1/f = 1/v + 1/u

1/10 = 1/v - 1/2v (substituting u = -2v)

1/10 = 2/2v - 1/2v

1/10 = 1/2v

10 = 2v

v = +5 cm ( + means it is right side of the mirror)

u = -2 * 5 = -10cm (- denotes that it is left side of the mirror)

Image distance = 5cm

Object distance = -10cm