Question

An object is placed at a distance of $15 cm$ in front of a convex mirror of radius of curvature $10 cm$
___________________
(a) Where will the image form ?
(b) Find the magnification m.
(c) What will be the nature of Image, real or virtual ?​

Answer 1

Explanation:

Here (u) distance of object from mirror= -10

Focal length= -(1/2 of radius of curvature)

= -(1/2* 12) =-6

So by mirror formula we get,

1/F=1/V+1/U

-1/6=1/V-1/10

1/V=1/10–1/6

1/V=3/30 -5/30

1/V= -2/30

V= -15

DISTANCE OF IMAGE FROM MIRROR IS 15 cm left side of mirror

Thus image will form 7.5 cm from the mirror

Answer 2

Given:

• R=10cm

• f=R2=5cm (positive)

• u= -15cm (negative)

• v=?

a. From relation: 1u+1v=1f

1v+1(−15)=15

⠀⠀⠀or

1v=15+115=415

⠀⠀⠀or

v=154=3.75cm

Thus the image will form at a distance 3.75 cm behind the mirror

b. Magnification m= −vu= −3.75(−15) = 14

Thus the size fo image is one fourth the size of the object.

c. The image will be virtual and erect.

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