Science, asked by Akash2504, 1 year ago

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

Answered by Anonymous
7

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.

Answered by doggogoddo41
0

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

Similar questions