Monochromatic light rays parallel to x-axis s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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):

Similar Questions

The minimum distance between an object and its real image formed by a convex lens of focal length $f$ is (in $f$)

$A$ screen is placed $90 \, cm$ from an object. The image of the object is formed on the screen by a convex lens at two different locations separated by $20 \, cm$. If the sizes of the images formed at these positions are $6 \, cm$ and $3 \, cm$,then the height of the object is.....$cm$.

$A$ lens is placed between a source of light and a wall. It forms images of area $A_1$ and $A_2$ on the wall for its two different positions. The area of the source of light is

Vedclass Test Series

Exam Paper Generator

Online Exam Module