Question

A convex mirror is used for real view on an automobile has a radius of curvature 2m. If a bus is located at a distance of 6m from the mirror. Find the position and nature of the image?​

Answer 1

Answer:

R=3m

f=R/2=3/2=1.5m

u=5m

v=?

Mirror formula:

1/f=1/u+1/v

by sign conventions:

1/1.5=1/-5+1/v

1/v=1/1.5+1/5⇒6.5/7.5

v=7.5/6.5

=1.15m

∴The image is virtual seen at the back of mirror.

m=hi/ho=-v/u

-1.15/-5=0.23

∴Image is erect and diminished in size by a factor of 0.23

Answer 2

Given :-

A convex mirror is used for real view on an automobile has a radius of curvature 2m.

A bus is located at a distance of 6m from the mirror.

To Find :-

The position and nature of the image.

Solution :-

We know that,

• f = Focal length
• r = Radius
• v = Distance of the image from the lens
• u = Distance of the object from the lens

By the formula,

$\underline{\boxed{\sf Focal \ length=\dfrac{Radius}{2} }}$

Given that,

Radius (r) = 2 m

Substituting their values,

$\sf =\dfrac{2}{2} =1 \ m$

Therefore, the focal length is 1 m.

By the formula,

$\underline{\boxed{\sf Mirror \ formula=\dfrac{1}{f} = \dfrac{1}{v} +\dfrac{1}{u} }}$

Substituting their values,

$\sf =\dfrac{1}{1} =\dfrac{1}{v} +\dfrac{1}{(-6)}$

$\sf =\dfrac{1}{1} +\dfrac{1}{6} =\dfrac{1}{v}$

$\sf =\dfrac{7}{6} =\dfrac{1}{v}$

$\sf =\dfrac{6}{7} =v$

Therefore, the position of the image is 6/7 m from the mirror.

By the formula,

$\underline{\boxed{\sf Magnification=\dfrac{-v}{u} }}$

By substituting them,

$\sf m=\dfrac{-(\dfrac{6}{7} )}{(-6)}$

$\sf m=\dfrac{1}{7}$

Hence, the magnification is 1/7.

Therefore,

Nature :- Virtual and erect.

Size :- Smaller than object.

Position :- 6/7 m behind the mirror.