The concave mirror has a radius of curvature of 3.0m. An object is placed in front of the mirror and an image with a magnification of 1.5 is formed. What can you deduce about the nature and location of the image? Draw a ray diagram to show the image formed by the mirror. Calculate the object and image distances.
Answers
Given:-
Radius of curvature = 3.0 m
Magnification = 1.5
To find:
Nature and position of the image.
The object and image distances.
Solution:-
Focal length = 3/2 = 1.5 m ...(i)
As,
We know, in a mirror,
-v/u = magnification
-v/u = 1.5
-v = 1.5u
v = -1.5u ...(ii)
Using mirror formula,
1/v + 1/u = 1/f
⇒ -1/1.5u + 1/u = 1/1.5 [∴From (i) & (ii)]
⇒ (-1 + 1.5)/1.5u = 1/1.5
⇒ 1.5u = 1.5(-1 + 1.5)
⇒ 1.5u = -1.5 + 2.25
⇒ 1.5u = 0.75
⇒ u = 0.75/1.5
⇒ u = 0.5 m
u = -0.5 m (as it is in the front of the mirror)
Now, putting the value of u = 0.5 m in eq.(ii):-
v = 1.5u
⇒v = 1.5*0.5
⇒v = 0.75 m
Thus, the object distance is 0.5 m and the image distance is 0.75 m.
For more help:
'u' is the object distance.
'v' is the image distance.
'f' is the focal length.
'R' is the radius of curvature.
Given:
R = -3 m (concave mirror)
m = 1.5 m
To find:
- Object and Image distance
- Natue and location of image
Solution:
f = R/2
=> f = -3/2 m
Let: u = -x m
Also, we know that by mirror formula:
1/v + 1/u = 1/f
=> 1/v + 1/-x = -2/3
=> 1/v = -2/3 + 1/x
=> 1/v = (3 - 2x)/3x
=> v = 3x/(3 - 2x)
Also, we know that:
m = -v/u
=> 1.5 = -v/u
=> 1.5u = -v
=> 1.5(-x) = -(3x/(3 - 2x))
=> -1.5x = -(3x/(3 - 2x))
=> 1.5 = 3/(3 - 2x)
=> 1.5(3 - 2x) = 3
=> 4.5 - 3x = 3
=> 1.5 = 3x
=> x = 0.5 m
u = -x = -0.5 m
v = 3x/(3 - 2x) = 1.5/(3 - 1) = 1.5/2 = 0.75 m
Nature and location:-
Nature:
Since u < f, the image formed will be virtual, larger than the object
Location:
Behind the mirror
________
Thus: