b) An object 5.0 cm in length is placed at distance of 20 cm in
front of a converging mirror of radius ofcurvature 30 cm.
position of the image, its nature and size.
Answers
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▶Here, Object distance, u = -20 cm (to the left of mirror)
Image distance, v = ? (To be calculated)
Radius of curvature, R = +30cm (It is a convex mirror)
So, Focal length,
Now, putting these values of u and applying formula:-
▶ 1/ V - 1/u = 1/f
1/V - 1/20 = 1/15
1/V = 1/15 + 1/20
1/V = 4+3 / 60
= 7/60
V = 60/7
V = +8.57 cm
Thus, the position of image is 8.57 cm behind the mirror (on its right side). Since the image is formed behind the convex mirror, therefore, the nature of image is virtual and erect.
Now,
For a mirror, Magnification,
m = - v /u
So, m = = + 0.42
Also, Magnification, m =
Where h2= Height of image, h1= Height of Object
So, +0.42 =
h2 = 0.42×5.0 = 2.1 cm
Thus, the size of image is 2.1 cm.
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1/ V - 1/u = 1/f
1/V - 1/20 = 1/15
1/V = 1/15 + 1/20
1/V = 4+3 / 60
= 7/60
V = 60/7
V = +8.57 cm
Since the image is formed behind the convex mirror, therefore, the nature of image is virtual and erect.
Now,
For a mirror, Magnification,
m = - v /u
So, m = = + 0.42
Also, Magnification, m =
Where h2= Height of image, h1= Height of Object
So, +0.42 =
h2 = 0.42×5.0 = 2.1 cm
Thus, the size of image is 2.1 cm.