(a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image, and also draw the ray diagram for above condition. (b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
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(a) h1 = 3 cm
u = -24 cm
f = 8 cm
Lens formula : 1/v – 1/u = 1/f
1/v – 1/-24 = 1/8
1/v = 1/12
V = 12 cm
Image is formed 12 cm behind the lens.
m = v/u = h2/h1
12/-24 = h2/3
h2 = -1.5 cm
Image is 1.5 cm high, real and inverted.
(b) u = - 3 cm
h1 = 3 cm
f = 8 cm
Lens formula : 1/v – 1/u = 1/f
1/v – 1/-3 = 1/8
1/v = - (5/24)
V = -4.8 cm
Image is formed 4.8 cm in front of the lens.
m = v/u = h2/h1
-4.8/-3 = h2/3
h2 = + 4.8 cm
Image is 4.8 cm high, virtual and erect.
hope this will help you...
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