science ch light class 10 example 10.1
Answers
Radius of mirror=+3.00
Object height= -5. 00
Image distance=?
height of the image=?
focal length=f=R/2=+3.00/3 =1.50m
1/v+1/u=1/f
or, 1/v=1/f -1/u
=+1/1.50 - 1/(-5.00)
=5.00+1.50/7.50
v=+7.50/6.50
=1.15m
the image is 1.15m at the back of the mirror.
magnification,m=h/h= -v/u = - 1.15/(-5.00)
=+0.23.
the image is virtual ,erect and smaller in size by 0.23.
Answer:
the image is virtual ,erect and smaller in size by 0.23.
Explanation:
Radius of mirror=+3.00
Object height= -5. 00
Image distance=?
height of the image=?
focal length=f=R/2=+3.00/3 =1.50m
1/v+1/u=1/f
or, 1/v=1/f -1/u
=+1/1.50 - 1/(-5.00)
=5.00+1.50/7.50
v=+7.50/6.50
=1.15m
the image is 1.15m at the back of the mirror.
magnification,m=h/h= -v/u = - 1.15/(-5.00)
=+0.23.
the image is virtual ,erect and smaller in size by 0.23.