Question

the radius of curvature of a convex mirror used as a rear view mirror in a moving car is '20' m a truck is coming from behind it at a distance of 3.5 metre calculate the a)position b)size of the image relative to the size of truck what will be the nature of the image.​

Correct answer please!!

Answer 1

Given:-

the radius of curvature of a convex mirror used as a rear view mirror in a moving car is '20' m a truck is coming from behind it at a distance of 3.5

Which means

• r = 20 m
• $f = \frac{r}{2} = \frac{20}{2} = 10m$
• Distance Of Object(u)= -3.5m

To Find:-

• position
• size of the image
• the nature of the image.

Solution:-

By Mirror Formula

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

$= > \frac{1}{10} = \frac{1}{v} + \frac{1}{ - 3.5}$

$= > \frac{1}{v} = \frac{1}{10} + \frac{1}{3.5}$

$= > \frac{1}{v} = \frac{1 \times 3.5}{10 \times 3.5} + \frac{1 \times 10}{3.5 \times 10}$

$= > \frac{1}{v} = \frac{3.5 + 10}{35}$

$= > \frac{1}{v} = \frac{13.5}{35}$

By Reciprocal

$= > v = \frac{35}{13.5}$

$= > v = 2.5m$

$\therefore \: magnification = \frac{ - v}{u}$

$= > m = \frac{ - 2.5}{ - 3.5}$

$= > m = 0.7$

Hence, The Size of the image formed will be 0.7 times the Size of Object.

As m is positive, so image formed is virtual and erect.

Answer 2

Answer:

f=r=2 = 2m +2=1m

u = -3.5 m

Using mirror formula,

1 ÷ -3.5 m +1 v 1 1m

1= v=1m +1 ÷ 3.5 m

1 v = (3.5+1) m = 3.5

1 v = 4.5 m = 3.5

1=v=9=7m

v = 7 - 9 m

v = 0.78 m

Image will be virtual, erect and smaller than

object