The formation of a real image using a biconve | vedclass

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(A) According to the lens maker's formula,the focal length $f$ of a lens in a medium is given by $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac

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Image disappears

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(A) According to the lens maker's formula,the focal length $f$ of a lens in a medium is given by $\frac{1}{f} = (\mu_{rel} - 1) (\frac{1}{R_1} - \frac{1}{R_2})$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.When the lens is immersed in water,the refractive index of the surrounding medium increases from $\mu_{air} \approx 1$ to $\mu_{water} \approx 1.33$. Consequently,the relative refractive index $\mu_{rel}$ decreases,which causes the focal length $f$ of the lens to increase.Since the focal length increases,the image position shifts away from the lens. Because the positions of the object and the screen are fixed,the image will no longer be formed on the screen. Therefore,the image will disappear from the screen.

The formation of a real image using a biconvex lens is shown below:
If the whole setup is immersed in water without disturbing the object and the screen positions,what will one observe on the screen?

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The formation of a real image using a biconvex lens is shown below:If the whole setup is immersed in water without disturbing the object and the screen positions,what will one observe on the screen?