Example 10.4 - Chapter 10 Class 10 - Light - Reflection and Refraction (Term 1)

Last updated at March 30, 2023 by Teachoo

A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.

A 2.0 cm tall object is placed perpendicular to the principal axis of a
convex lens of focal length 10 cm. The distance of the object from the
lens is 15 cm. Find the nature, position and size of the image. Also
find its magnification.
Object is always placed above principal axis.
Hence, height of object is always positive.
Height of object = h = + 2 cm
Focal length of a convex lens is positive
Focal length = f = +10 cm
Object is always placed infront of the lens
Hence, object distance is always negative
Object distance = u = β 15 cm
Finding image distance
Let the image distance = v
Using lens formula
1/π = 1/π£ β 1/π’
1/π + 1/π’ = 1/π£
1/π£ = 1/π + 1/π’
1/π£ = 1/10 + 1/((β15))
1/π£ = 1/10 β 1/15
1/π£ = (3 β 2)/30
1/π£ = 1/30
π£ = +30 cm
Therefore,
Image is at a distance of 30 cm from the lens
Positive sign indicates that image is formed behind the lens
β΄ Image is real
Finding size of the image
Let the height of the image = hβ
We know that
Magnification by a lens = π£/π’ = (πΌππππ πππ π‘ππππ)/(ππππππ‘ πππ π‘ππππ)
Also,
Magnification by a lens = β^β²/β = (π»πππβπ‘ ππ πππππ)/(π»πππβπ‘ ππ ππππππ‘)
From (1) and (2)
π£/π’ = β^β²/β
30/(β15) = β^β²/2
β^β²/2 = (β30)/15
hβ = 2 Γ β2
hβ = β4 cm
Therefore, Height of image is 4 cm
Negative sign indicated that image is formed below the principal axis
β΄ Image is inverted
Finding magnification
Magnification of a lens = β^β²/β
= (β4)/2
= β2
β΄ The image is two times enlarged
Hence,
Image is real, inverted and enlarged by two times

CA Maninder Singh is a Chartered Accountant for the past 13 years and a teacher from the past 17 years. He teaches Science, Economics, Accounting and English at Teachoo

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