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CBSE Class 12 Physics 2019 Delhi Set 1 Paper

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Question : 7 of 27
Marks: +1, -0
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 −5 D-5\,\text{D} ?
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 425\frac{4\sqrt{2}}{5}.
Solution:  
Calculation of focal length
Lens maker's formula
Calculation of radius of curvature
f=1P=1−5 m=−1005 cm=−20 cmf = \frac{1}{P} = \frac{1}{-5}\,\text{m} = -\frac{100}{5}\,\text{cm} = -20\,\text{cm}
1f=(μ2μ1−1)(1R1−1R2)\frac{1}{f} = \left( \frac{\mu_2}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
μ2=1.5, μ1=1.4, R1=−R, R2=R\mu_2 = 1.5,\ \mu_1 = 1.4,\ R_1 = -R,\ R_2 = R
1−20=(1.51.4−1)(−1R−1R)\frac{1}{-20} = \left( \frac{1.5}{1.4} - 1 \right) \left( -\frac{1}{R} - \frac{1}{R} \right)
1−20=(0.11.4)(−2R)\frac{1}{-20} = \left( \frac{0.1}{1.4} \right) \left( -\frac{2}{R} \right)
R=207 cm (=2.86 cm)R = \frac{20}{7}\,\text{cm}\,(=2.86\,\text{cm})
OR
Formula
Substitution and calculation
μ=sin⁡(A+δm2)sin⁡(A2)\mu = \frac{\sin\left( \frac{A+\delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)}
μ=μ1μ2=1.6452=842=2\mu = \frac{\mu_1}{\mu_2} = \frac{1.6}{\frac{4}{5}\sqrt{2}} = \frac{8}{4\sqrt{2}} = \sqrt{2}
μ=μ1μ2=1.6452=842=2\mu = \frac{\mu_1}{\mu_2} = \frac{1.6}{\frac{4}{5}\sqrt{2}} = \frac{8}{4\sqrt{2}} = \sqrt{2}
2=sin⁡(60∘+δm2)sin⁡(60∘2)\sqrt{2} = \frac{\sin\left( \frac{60^{\circ} + \delta_m}{2} \right)}{\sin\left( \frac{60^{\circ}}{2} \right)} =sin⁡(60∘+δm2)sin⁡30∘= \frac{\sin\left( \frac{60^{\circ} + \delta_m}{2} \right)}{\sin 30^{\circ}}
∴sin⁡(60+δm2)=2⋅12\therefore \sin\left( \frac{60+\delta_m}{2} \right) = \sqrt{2} \cdot \frac{1}{2} =12=sin⁡45∘= \frac{1}{\sqrt{2}} = \sin 45^{\circ}
∴60+δm2=45∘\therefore \frac{60+\delta_m}{2} = 45^{\circ}
∴δm=30∘\therefore \delta_m = 30^{\circ}
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