7. An erect image 5 times the size of the object is obtained with a concave mirror of
radius of curvature 40cm. What is the position of the object?
Answers
Answer:
Explanation:
magnification m = - v/u = h'/h = +3
v = -3 u actually u is negative. Hence v is positive.
R = -36 cm => f = R/2 = -18 cm , negative for a concave mirror.
1/v + 1/u = 1/f, mirror equation
1/(-3u) + 1/u = 1/(-18)
2/(3u) = -1/18
u = - 12 cm v = +36 cm
the object is placed 12 cm from pole of concave mirror , less than the focal length, to have an ERECT magnified image 3 times the object.
Answer:
in order to find the position, we could use the object distance and there two scenarios where m=-5(or) m=+5
Explanation:
R=2f
R/2=f =
40/2=f=20
take U to be object distance , take V to be image distance
magnification be M
if m=-v/u=+5 if m=-v/u=-5
v=-5u v=5u
1/f=1/v+1/u 1/f=1/v+1/u
1/20=1/-5u+1/u 1/20=1/5u+1/u
1/20=-5+1/-5u 1/20=5+1/5u
1/20=-4/-5u 5u= 120
-5u=-80 u= 24
u=16 v=-5u =-80 v= 5u =120
in bot cases object is positive hence it is behind the mirror