An object is placed in front of convex lens made of glass. How does th

1.	An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?
Answer» From the lens maker formula, it is clear that n₂₁ decreases then focal length increase. \(\frac {1}{f} = (n_{21} - 1)(\frac{1}{r_1}-\frac{1}{r_2}) (n_{21} =\frac {n_2}{n_1})\) Here refractive index of the glass with respect to surrounding material decreases. Hence, focal length increases which will also increase the image distance.

An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?

Answer»

From the lens maker formula, it is clear that n₂₁ decreases then focal length increase.

\(\frac {1}{f} = (n_{21} - 1)(\frac{1}{r_1}-\frac{1}{r_2}) (n_{21} =\frac {n_2}{n_1})\)

Here refractive index of the glass with respect to surrounding material decreases. Hence, focal length increases which will also increase the image distance.

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An object is placed in front of convex lens made of glass. How does the image distance vary if the refractive index of the medium is increased in such a way that still it remains less than the glass?

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