A U-shaped wire is placed before a concave mirror having radius of curvature 20cm as shown in figure (18-E1). Find the total length of the image.
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Given :
R= 20cm
f= R/2 = 20/2 = 10cm
For part AB, PB =u= 30+ 10 = 40cm
so u= - 40cm
1/v+1/u=/1f
1/v= 1/f- 1/u
= -1/10 + 1/40
= - 4+1/40
v= - 40/3 = -1 3.3cm=
m= -v/u = -(-13.3)/ -40 = -1/3
hi/ho= m
hi= mx ho=-10/3 = - 3.33cm
For part CD, Pc= 30
u= -30cm
1/v=1/f-1/u = 1/10 +1/30 = - 2/30
v= - 15cm= Pc'
m= C1D1/CD = - v/u = -(-15)/30 = -1/2
C1D1= 5cm
B1C1=Pc1- Pb1= 15- 13.3 = 1.7cm
so total length is A1B1+B1C1+ C1D1= 3.3 +1.7 +5= 10cm
R= 20cm
f= R/2 = 20/2 = 10cm
For part AB, PB =u= 30+ 10 = 40cm
so u= - 40cm
1/v+1/u=/1f
1/v= 1/f- 1/u
= -1/10 + 1/40
= - 4+1/40
v= - 40/3 = -1 3.3cm=
m= -v/u = -(-13.3)/ -40 = -1/3
hi/ho= m
hi= mx ho=-10/3 = - 3.33cm
For part CD, Pc= 30
u= -30cm
1/v=1/f-1/u = 1/10 +1/30 = - 2/30
v= - 15cm= Pc'
m= C1D1/CD = - v/u = -(-15)/30 = -1/2
C1D1= 5cm
B1C1=Pc1- Pb1= 15- 13.3 = 1.7cm
so total length is A1B1+B1C1+ C1D1= 3.3 +1.7 +5= 10cm
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