NCERT Question 14 - Chapter 10 Class 10 - Light - Reflection and Refraction
Last updated at May 29, 2023 by Teachoo
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex
mirror of radius of curvature 30 cm. Find the position of the image, its nature
and size.
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NCERT Question 14
An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Since object is always placed above the principal axis
Height of the object is always positive
Height of the object = h = + 5 cm
Since the object is placed in front of the mirror
Object distance will be negative
Object distance = u = −20 cm
Given
Radius of curvature = 30 cm
We know that,
Focal length = (𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒)/2
= 30/2
= 15 cm
Since focus of a convex mirror is on the right side
The focal length will be positive
Focal length = f = + 15 cm
Finding position of the image
Let the image distance = v
Using mirror formula
1/𝑓 = 1/𝑢 + 1/𝑣
1/𝑓 − 1/𝑢 = 1/𝑣
1/𝑣 = 1/𝑓 − 1/𝑢
1/𝑣 = 1/15 − 1/((−20))
1/𝑣 = 1/15 + 1/20
1/𝑣 = (4 + 3)/60
1/𝑣 = 7/60
𝑣 = 60/7 cm
𝑣 = + 8.6 cm
Thus, Image is produced at a distance of 8.6 cm
Positive sign denotes the image is formed behind the mirror.
Hence, the image is erect
Finding size of image
Let the height of the image = h’
We know that,
Magnification of a mirror = (−(𝐼𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒))/(𝑂𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)
= 𝑣/𝑢
Also,
Magnification of a mirror = (𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑂𝑏𝑗𝑒𝑐𝑡)/(𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑚𝑎𝑔𝑒)
= ℎ^′/ℎ
From (1) and (2),
(−𝑣)/𝑢 = ℎ^′/ℎ
(−(60/7))/((−20)) = ℎ^′/5
60/(7 × 20) = ℎ^′/5
ℎ^′/5 = 6/(7 × 2)
ℎ^′ = (6 × 5)/(7 × 2)
ℎ^′ = (3 × 5)/7
ℎ^′ = 15/7
ℎ^′ = 2.2 cm
Since height of object is 5 cm
∴ Image produced is diminished
And since height is positive,
image is erect
Therefore,
Image is virtual, erect and diminished with size 2.2 cm
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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