SATHEE: Optics Question 680

Optics Question 680

Question: A point object is placed at the center of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is

[IIT-JEE (Screening) 2004]

Options:

A) 2 cm

B) 4 cm

C) 6 cm

D) 12 cm

Show Answer

Answer:

Correct Answer: C

Solution:

Using refraction formula $\frac{_{1} μ_{2} - 1}{R} = \frac{_{1} μ_{2}}{v} - \frac{1}{u}$ in given case, medium (1) is glass and (2) is air .

So $\frac{_{g} μ_{a} - 1}{R} = \frac{_{g} μ_{a}}{v} - \frac{1}{u}$

$\Rightarrow \frac{\frac{1}{1.5} - 1}{- 6} = \frac{1}{1.5 v} - \frac{1}{- 6}$

$\Rightarrow \frac{1 - 1.5}{- 6} = \frac{1}{v} + \frac{1.5}{6}$

$\Rightarrow \frac{0.5}{6} = \frac{1}{v} + \frac{1}{4}$

$\Rightarrow \frac{1}{v} = \frac{1}{12} - \frac{1}{4} = - \frac{2}{12} = - \frac{1}{6}$

Therefore v = 6 cm

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Optics Question 680

Question: A point object is placed at the center of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the virtual image from the surface of the sphere is

Options:

Answer:

Solution:

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