SATHEE: Optics 3 Question 21

Optics 3 Question 21

21. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature $R$ . On immersion in a medium of refractive index 1.75 , it will behave as a

(a) convergent lens of focal length $3.5 R$

$(1999, 2 M)$

(b) convergent lens of focal length $3.0 R$

(d) divergent lens of focal length $3.0 R$

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Answer:

Correct Answer: 21. (a)

Solution:

$R_{1} = - R, R_{2} = + R, μ_{g} = 1.5$ and $μ_{m} = 1.75$

$∴ \frac{1}{f} = \frac{μ_{g}}{μ_{m}} - 1 \frac{1}{R_{1}} - \frac{1}{R_{2}}$

Substituting the values, we have

$\begin{aligned} \frac{1}{f} & = \frac{1.5}{1.75} - 1 \frac{1}{- R} - \frac{1}{R} = \frac{1}{3.5 R} \\ ∴ & f & = + 3.5 R \end{aligned}$

Therefore, in the medium it will behave like a convergent lens of focal length 3.5R. It can be understood as, $μ_{m} > μ_{g}$ , the lens will change its behaviour.

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Optics 3 Question 21

21. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature $R$ . On immersion in a medium of refractive index 1.75 , it will behave as a

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics 3 Question 21

21. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature R. On immersion in a medium of refractive index 1.75 , it will behave as a

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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21. A concave lens of glass, refractive index 1.5 has both surfaces of same radius of curvature $R$ . On immersion in a medium of refractive index 1.75 , it will behave as a