CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 7
Total: 27
Calculate the radius of curvature of a equiconcave lens of refractive index 1.5 , when it is kept in a medium of refractive index 1.4, to have a power of 5D ?
OR
An equilateral glass prism has a refractive index 1.6 in air. Calculate the angle of the minimum deviation of the prism, when kept in a medium of refractive index
42
5
.
Solution:  
Calculation of focal length
Lens maker's formula
Calculation of radius of curvature
f=
1
P
=
1
5
m
=
100
5
cm
=20cm

1
f
=(
µ2
µ1
1
)
(
1
R1
1
R2
)

µ2=1.5,µ1=1.4,R1=R,R2=R
1
20
=(
1.5
1.4
1
)
(
1
R
1
R
)

1
20
=(
0.1
1.4
)
(
2
R
)

R=
20
7
cm
(=2.86cm)

OR
Formula
Substitution and calculation
µ=
sin
(A+δm)
2
sin
A
2

µ=
µ1
µ2
=
1.6
4
5
2
=
8
42
=2

µ=
µ1
µ2
=
1.6
4
5
2
=
8
42
=2

2=
sin(
60+δm
2
)
sin
60
2
=
sin(
60+δm
2
)
sin30

sin(
60+δm
2
)
=2
1
2
=
1
2
=sin45

60+δm
2
=45

δm=30
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