solve numerical based on R and F(7 numericals ),mirror formula( 18 numericals)
Answers
Answer:
A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?
Soluion : V=-50 cm F=-20cm
From mirror formula 1/u = 1/f – 1/v
= -1/20+ 1/50 =-3/100 U = -33.3 cm
Therefore the distance of the object from the wall x = 50 – u
X = 50 – 33.3 = 16.7 cm.
Q. 2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced?
Answer: Here f = -15 cm, u = -40 cm
Now 1/f = 1/u + 1/v
Then 1/v = 1/f – 1/u
Or V= uf/u-f = -40 X -15/25 = -24 cm
Then object is displaced towards the mirror let u1 be the distance object from the
Mirror in its new position.
Then u1 = -(40-20) = -20cm
If the image is formed at a distance u1 from the mirror then
v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 cm.
Therefore the image will move away from the concave mirror through a distance equal to 60 – 24 = 36 cm.
Numerical based on mirror formula :---
Q.1. the object is set up .5 m away from a converging mirror. If the focal length of the lens is .2 m, determine a) the location of the real image, and b) the magnification of the image.
Ans. a) In order to determine the location of the image, we'll use the Mirror Formula.
1f=1v+1uu=−0.5 m,f=−0.2 m1v=−10.2+10.5v=−0.33 m
Here “–” sign shows that image is real.
b) Now that we have the location of the image, we can find the magnification.
m=−vu=−(−0.33)(−0.5)=−0.66
As
|m|<1
, the size of image is smaller than the object and negative sign shows that image is real and inverted.
Q.2. An object of height 10.0 cm is positioned 30.0 cm from a concave mirror with a focal length 8.0 cm.
Find the image position.
Solution:
We are given
ho=10.0 cm,u=−30 cm
and
f=−8 cm
Using Mirror equation
1f1vv=1v+1u=1f−1u=−10.9 cm
Since the image distance is negative, the image is real.