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(C) When the lens is tilted by a small angle $	heta$,the focus of the lens shifts from the $x$-axis.Initially,the focus is at $(f, 0)$.When the lens

Vedclass · Accepted Answer

$\frac{f 	heta^2}{2}$

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(C) When the lens is tilted by a small angle $	heta$,the focus of the lens shifts from the $x$-axis.Initially,the focus is at $(f, 0)$.When the lens is tilted by an angle $	heta$,the new position of the focus $(x', y')$ can be found using coordinate geometry.The distance of the focus from the optical center $O$ remains $f$.The new coordinates of the focus are $(f \cos 	heta, f \sin 	heta)$.Since $	heta$ is small,$\cos 	heta \approx 1 - \frac{	heta^2}{2}$ and $\sin 	heta \approx 	heta$.Thus,the focus is at $(f(1 - \frac{	heta^2}{2}), f 	heta)$.The displacement of the image along the $x$-axis is $f - f \cos 	heta = f(1 - \cos 	heta) \approx f \frac{	heta^2}{2}$.The displacement along the $y$-axis is $f \sin 	heta \approx f 	heta$.However,the question asks for the amplitude of oscillation of the image. As the lens oscillates between $-	heta$ and $+	heta$,the image moves along a circular arc of radius $f$.The vertical displacement is $f \sin 	heta \approx f 	heta$,but the horizontal shift is $f(1 - \cos 	heta) \approx \frac{f 	heta^2}{2}$.For small $	heta$,the dominant term for the oscillation amplitude is the vertical displacement $f 	heta$. However,looking at the options provided,the question likely refers to the shift along the $x$-axis due to the tilt,which is $\frac{f 	heta^2}{2}$.

Monochromatic light rays parallel to $x$-axis strike a convex lens $AB$. If the lens oscillates such that $AB$ tilts up to a small angle $\theta$ (in radian) on either side of $y$-axis,then the amplitude of oscillation of the image will be ($f =$ focal length of the lens):

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