Example 10.1 - Chapter 10 Class 10 - Light - Reflection and Refraction
Last updated at Sept. 10, 2019 by Teachoo
A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image
Example 10.1
A convex mirror used for rear-view on an automobile has a radius of
curvature of 3.00 m. If a bus is located at 5.00 m from this mirror,
find the position, nature and size of the image
Radius of curvature = 3 m
We know that
Focal length = (π ππππ’π ππ ππ’ππ£ππ‘π’ππ)/2
= 3/2
= 1.5 m
Since the focus of convex mirror is behind the mirror, the focal length will be positive
Focal length = f = + 1.5 m
Now,
Bus (Object) is placed infront of the mirror
β΄ Object distance will be negative
Object distance = u = β5 m
Finding position of image
Let the image distance = v
Using mirror formula,
1/π = 1/π£ + 1/π’
1/π β 1/π’ = 1/π£
1/π£ = 1/π β 1/π’
1/π£ = 1/1.5 β 1/((β5))
1/π£ = 1/1.5 + 1/5
1/π£ = 10/15 + 1/5
1/π£ = (10 + 3)/15
1/π£ = 13/15
1/π£ = 15/13
π£ = + 1.15 m
β΄ Image is produced at a distance of 1.15 m
+ve sign indicates that image is formed behind the mirror.
Finding size of image
Magnification = (βπ£)/π’
= (β (πΌππππ πππ π‘ππππ))/((ππππππ‘ πππ π‘ππππ))
= (β(1.15))/(β5)
= 1.15/5
= 115/500
= 23/100
= + 0.23
Since the magnification is less than 1
β΄ Image is diminished
and it is smaller in size by a factor of 0.23
And since magnification is positive,
β΄ Image is erect
Hence,
Image is virtual, erect and smaller in size by a factor of 0.23
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.