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CBSE Class 12 Physics 2016 Outside Delhi Set 1 Paper

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Question : 25 of 26
Marks: +1, -0
(i) Derive the mathematical relation between refractive indices n1n_1 and n2n_2 of two radii and radius of curvature RR for refraction at a convex spherical surface. Consider the object to be a point since lying on the principle axis in rarer medium of refractive index n1n_1 and a real image formed in the denser medium of refractive index n2n_2. Hence, derive lens maker's formula.
(ii) Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm20\ \mathrm{cm}. The distance of light source from the glass surface is DD. At what position is the image formed?
OR
(a) Draw a labelled ray diagram to obtain the real image formed by an astronomical telescope in normal adjustment position. Define its magnifying power.
(b) You are given three lenses of power 0.5 D, 4 10 D10\ \mathrm{D} and n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} to design a telescope.
(i) Which lenses should be used as objective and eyepiece? Justify your answer.
(ii) Why is the aperture of the objective preferred to be large? .
Solution:  
(i) Derivation of 1f=(n2−nn1)(1R1−1R2)\frac{1}{f} = \left(\frac{n_2 - n}{n_1}\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)
θ1=α+β\theta_1 = \alpha + \beta
Ray diagram showing real image formation as per description
∴ θ2=β−γ\therefore\ \theta_2 = \beta - \gamma
γ=β−θ\gamma = \beta - \theta
θ1\theta_1
For paraxial rays θ2\theta_2 and n2sin⁡θ2=n1sin⁡θ2n_2 \sin \theta_2 = n_1 \sin \theta_2 are small
Therefore, sin⁡isin⁡r≈ir=n2n1\frac{\sin i}{\sin r} \approx \frac{i}{r} = \frac{n_2}{n_1} (Snell's law)
Reduces to
∴i×n1=r×n2\therefore i \times n_1 = r \times n_2
(α+β)n1=(β−γ)n2(\alpha + \beta) n_1 = (\beta - \gamma) n_2
n1(NMOM+NMMC)=(NMMC−NMMI)n2n_1 \left( \frac{NM}{OM} + \frac{NM}{MC} \right) = \left( \frac{NM}{MC} - \frac{NM}{MI} \right) n_2
n2(1−u+1+R)=(1+R−1u)n2n_2 \left( \frac{1}{-u} + \frac{1}{+R} \right) = \left( \frac{1}{+R} - \frac{1}{u} \right) n_2
n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}
n2−n1R1=n2v′−n1u\frac{n_2 - n_1}{R_1} = \frac{n_2}{v'} - \frac{n_1}{u}
Applying above relations to refraction through a lens:
For surface 1
n1−n2R2=n1v−n2v′\frac{n_1 - n_2}{R_2} = \frac{n_1}{v} - \frac{n_2}{v'} ......(i)
For surface 2
(n2−n1)(1R1−1R2)=n1(1v−1u)(n_2 - n_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = n_1 \left( \frac{1}{v} - \frac{1}{u} \right) ......(ii)
Adding eqn. (i) and (ii),
u∝v=fu \propto v = f
For ∴ n1f=(n2−n1)(1R1−1R2)\therefore\ \frac{n_1}{f} = (n_2 - n_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
⇒ 1f=(n2n1−1)(1R1−1R2)\Rightarrow\ \frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
R=20 cm, n2=1.5, n1=1, u=−100 cmR = 20\ \mathrm{cm},\ n_2 = 1.5,\ n_1 = 1,\ u = -100\ \mathrm{cm}
(iii) n2v=n2−n1R+n1u\frac{n_2}{v} = \frac{n_2 - n_1}{R} + \frac{n_1}{u}
=0.520 cm−1100 cm= \frac{0.5}{20\ \mathrm{cm}} - \frac{1}{100\ \mathrm{cm}}
=1.5100 cm= \frac{1.5}{100}\ \mathrm{cm}
⇒v=100 cm\Rightarrow v = 100\ \mathrm{cm}
100 cm100\ \mathrm{cm} a real image on the other side, =5 D= 5\ \mathrm{D} away from the surface.
OR
(a) Labelled ray diagram of Astronomical Telescope
Definition of magnifying power
(b) (i) Identification of lenses Justification
(ii) Reason
Definition-It is the ratio of the angle subtended at the eye, by the final image, to the angle which the object subtends at the lens, or the eye.
(b) (i) Objective =10 D= 10\ \mathrm{D}
Eye lens M=f0fe=PeP0M = \frac{f_0}{f_e} = \frac{P_e}{P_0}
This choice would give higher magnification as
M=  f0/fe=  Pe/P0{M}=\;{f_0}/{f_e}=\;{P_e}/{P_0}
(ii) High resolving power / Brighter image / lower limit of resolution
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