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(D) Given: Separation between the two positions of the lens,$d = 10\, cm$.Ratio of the sizes of the images in the two positions,$\frac{I_1}{I_2} = \fr

Vedclass · Accepted Answer

$99$

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(D) Given: Separation between the two positions of the lens,$d = 10\, cm$.Ratio of the sizes of the images in the two positions,$\frac{I_1}{I_2} = \frac{3}{2}$.Let $D$ be the distance between the object and the screen.Using the displacement method formula for a thin lens,the ratio of the image sizes is given by $\frac{I_1}{I_2} = \frac{(D+d)^2}{(D-d)^2}$.Substituting the given values: $\frac{3}{2} = \frac{(D+10)^2}{(D-10)^2}$.Taking the square root on both sides: $\sqrt{\frac{3}{2}} = \frac{D+10}{D-10}$.This leads to $\sqrt{3}(D-10) = \sqrt{2}(D+10)$.$D(\sqrt{3} - \sqrt{2}) = 10(\sqrt{3} + \sqrt{2})$.$D = 10 	imes \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = 10 	imes (\sqrt{3} + \sqrt{2})^2 = 10 	imes (3 + 2 + 2\sqrt{6}) = 10(5 + 2\sqrt{6}) \approx 10(5 + 4.899) = 98.99\, cm \approx 99\, cm$.

An object is located in a fixed position in front of a screen. $A$ sharp image is obtained on the screen for two positions of a thin lens separated by $10\, cm$. The size of the images in the two situations are in the ratio $3 : 2$. What is the distance between the screen and the object in $cm$?

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