Question

Read the following and answer any four:
Convex mirrors are used as rear view mirror in vehicles. The image formed by the
convex mirror is diminished due to which it gives wider field of view of the traffic
behind the vehicle. Consider the convex mirror used on a moving vehicle has radius of
curvature 2m and a truck is coming behind it by maintaining a constant distance of 3.5
m.
(a) The distance behind the mirror where the image is formed is
(i) 0.28 m (ii) 1.5 m (iii) 0.78m (iv) 7.8 m
(b) The nature of image formed is
(i) virtual and erect (ii) real and inverted (iii) real erect and enlarged
(iv) none of these
(c) The size of the image of the truck is
(i).30 (ii) .58 (iii) .78 (iv).22
(d) The focal length of the mirror
(i) 0.5m (ii) lm (iii) 1.5 m (iv) 2m
(e) If instead of 3.5 m ,truck maintain a distance of 2m, the image formed will be
(i) real, erect and diminished
(ii) virtual , inverted and diminished
(iii)real, erect and enlarged
(iv) virtual erect, diminished

Answer 1

Answer:

four answers are given as :-

(b)(i) virtual and erect

(c)(iv)0.22

(d)(ii) 1m

(e)(iv) virtual, erect and diminished

Explanation:

The distance of object(truck) from mirror is u and the distance of image formed from the mirror is v.

Distance of truck from the mirror, u = -3.5m

Radius of curvature, R = 2m

R = 2f where f is the focal length of the mirror

f = R/2 ⇒ f = 1m

Using the mirror formula , $\frac{1}{f}=\frac{1}{v} +\frac{1}{u}$

Putting the values of f and u in this formula, we get

$1 = \frac{1}{v} +\frac{1}{-3.5}$

$\frac{1}{v} =1+\frac{1}{3.5}$

(b)therefore, v = 0.78m (virtual and erect)

Magnification (m) = $\frac{h_{2} }{h_{1} } =\frac{-v}{u}$

where, $h_{2}$ = size of image

$h_{1}$ = size of object

$\frac{h_{2} }{h_{1} }=-\frac{0.78}{-3.5}$

$h_{2} = 0.22h_{1}$

(c)Hence, the size of the image relative to the size of the truck is  0.22

(d) focal length = 1m

(e) now u = 2m

using mirrors formula $\frac{1}{f} =\frac{1}{v}+\frac{1}{u}$

$1 =\frac{1}{v} +\frac{1}{-2}$

$\frac{1}{v} = 1 +\frac{1}{2}$

$\frac{1}{v} = \frac{3}{2}$

$v=\frac{2}{3}$

v = 0.6m

therefore the image formed will be virtual , erect and diminished