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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 ‌
4√2
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
4√2
=√2
µ‌=‌
µ1
µ2
=‌
1.6
4
5
√2
=‌
8
4√2
=√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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