Question

A convex mirror of radius of curvature 1.6 cm has an object placed at distance of 1 metre from it the image and the magnification

Answer 1

f= -0.8= \dfrac{4}{5}f=−0.8=

5

4

; u= -1u=−1

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

v

1

+

u

1

=

f

1

\dfrac{1}{v}-\dfrac{1}{1}=\dfrac{5}{4}

v

1

−

1

1

=

4

5

v=\dfrac{4}{9}v=

9

4

m

v>0v>0 so it is formed behind the mirror.

Answer 2

The distance of the image is 0.007 meters and the magnification of image is 143.

GIVEN

The radius of curvature = 1.6cm

Object distance = 1 meter

TO FIND

Image distance

Magnification of the image formed.

SOLUTION

We can simply solve the above problem as under-

To calculate the image distance and magnification we need to find the focal length of the convex lens

Focal length is the half of Radius of curvature

So,

$f = \frac{1.6}{2} = 0.8cm$

or, f = 0.008 meters

We can find out the image distance by applying the mirror formula -

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

where,

f = focal length of the mirror = 0.0008 m

u = object distance = -1 meters

v = image distance

Putting the values in the above formula, we get

$\frac{1}{0.008} = \frac{1}{v} - 1$

$\frac{1}{v} = \frac{1}{0.008} + 1$

$\frac{1}{v} = 126$

$v = \frac{1}{126} = 0.007$

Image distance = 0.007 meter

Now,

Magnification

$= \frac{u}{v}$

where,

u = object height

v = image height

putting the values in the above formula,

Magnification = 1/0.007 =

143.

So, Magnification is 143.

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