Question:
Formation of real image using a biconvex lens is shown below :
If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen ?
Formation of real image using a biconvex lens is shown below :
If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen ?
If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen ?
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Lens maker formula: in water the focal length of a glass lens increases to about four times. Compare it with the object distance (between f and 2f in air).
Updated On: Nov 5, 2026
- Image disappears
- Magnified image
- Erect real image
- No change
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The Correct Option is A
Solution and Explanation
In air, the object lies between f and 2f of the lens (as in the figure) and a real, inverted, magnified image forms on the screen beyond 2f.
Lens maker formula: \(\frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)\). In air \((n_m=1)\): factor \(=0.5\); in water \((n_m\approx1.33)\): factor \(=\frac{1.5}{1.33}-1\approx0.13\).
So \(f_w\approx\frac{0.5}{0.13}f\approx4f\).
The object distance (less than 2f) is now less than \(f_w\), so the object lies inside the focal length and the lens forms a virtual image. No real image can fall on the screen: the image disappears from the screen.
Answer: (1) Image disappears
Lens maker formula: \(\frac1f=\left(\frac{n_g}{n_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)\). In air \((n_m=1)\): factor \(=0.5\); in water \((n_m\approx1.33)\): factor \(=\frac{1.5}{1.33}-1\approx0.13\).
So \(f_w\approx\frac{0.5}{0.13}f\approx4f\).
The object distance (less than 2f) is now less than \(f_w\), so the object lies inside the focal length and the lens forms a virtual image. No real image can fall on the screen: the image disappears from the screen.
Answer: (1) Image disappears
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