Question

A convex mirror with a radius of curvature of 3m is used as rear view in an automobile. If a busis located at 5m from this mirror, find the position, nature and size of the imaze.

Answer 1

Dear Student,

# Given -

r = 3 m

u = -5 m

# Solution -

Focal length of mirror is half of its radius of curvature.

f = r/2

f = 1.5 m

Now, using mirror formula -

1/f = 1/u + 1/v

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

1/v = 1/1.5 + 1/5

1/v = (5+1.5) / (5×1.5)

v = 7.5 / 6.5

v = 1.154 m

Magnification is calculated by -

M = -v/u

M = -1.154/-5

M = 0.231

Therefore, image is formed at 1.154 m behind the mirror and it will be virtual, erect and diminished (M<1).

Hope this helps you...

Answer 2

Answer:

Position = 1.154 m.

Nature = virtual, erect and diminished.

Size = 0.23 times the size of the bus.

Explanation:

Sign Convention:

All distances which are measured along the direction of incident of light are taken as positive and those which measured along direction opposite to that of incident of light are taken as negative.

Given:

Radius of curvature of mirror, R = 3 m.

Distance of bus from the mirror = -5 m.

The focal length of the mirror, $f = \dfrac R2= \dfrac 32 = 1.5 \ m.$

The position of image is obtained using the mirror formula as,

$\dfrac 1f = \dfrac 1 v+ \dfrac 1u \\\dfrac 1v = \dfrac 1f - \dfrac 1u \\=\dfrac {1}{1.5} - \dfrac{1}{-5}\\=\dfrac {1}{1.5} +\dfrac{1}{5}\\=\dfrac{5+1.5}{5\times 1.5}\\=\dfrac{6.5}{7.5}\\\Rightarrow v = \dfrac{7.5}{6.5} = 1.154\ m.$

Also, $v$ is positive, means the image is formed behind the mirror i.e., virtual in nature.

The magnification of the bus is given as

$m = \dfrac {-v}{u}\\= \dfrac{-1.154}{-5}\\=0.23.$

$m$ is positive means the image is upright (or erect) and size of the image of the bus is 0.23 times the size of the bus.