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An object is placed in front of a 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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Answer» From lens makers formula, \(\frac{1}{f}\) = \((\frac{μ_g}{μ _m}-1)\) \((\frac{1}{R_1}-\frac{1}{R_2})\) When µm is increased but still less than µg, Means \((\frac{μ_g}{μ _m}-1)\) decreases. It means, \(\frac{1}{f}\) also decreases. Now, From \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\), When \(\frac{1}{f}\) decreases, \(\frac{1}{v}\) also decreases, Therefore, V increases that is distance of image increases away from lens. |
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