A point source of light is moving at a rate o | vedclass

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

Question

(B) From the lens formula,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad \dots(i)$Here,$f = +10 \,cm$ and $u = -15 \,cm$.Using the lens formul

Vedclass · Accepted Answer

$8 \,cm \,s^{-1}$ towards the lens

Vedclass · Answer

(B) From the lens formula,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \quad \dots(i)$Here,$f = +10 \,cm$ and $u = -15 \,cm$.Using the lens formula: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$.Thus,$v = +30 \,cm$.Differentiating equation $(i)$ with respect to time $t$:$-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$$\frac{dv}{dt} = \left( \frac{v^2}{u^2} ight) \frac{du}{dt}$Given that the source is moving towards the lens,$\frac{du}{dt} = -2 \,cm \,s^{-1}$ (since $u$ is becoming less negative).$\frac{dv}{dt} = \left( \frac{30}{-15} ight)^2 	imes (-2 \,cm \,s^{-1}) = (2)^2 	imes (-2 \,cm \,s^{-1}) = 4 	imes (-2) = -8 \,cm \,s^{-1}$.The negative sign indicates that the image is moving in the direction of decreasing $v$ (towards the lens). However,since the image is formed on the other side $(v = +30)$,a negative rate of change for $v$ means the image is moving towards the lens. Wait,let's re-evaluate: If $v$ is positive and $\frac{dv}{dt}$ is negative,the distance from the lens is decreasing. Therefore,the image is moving towards the lens at $8 \,cm \,s^{-1}$.

$A$ point source of light is moving at a rate of $2 \,cm \,s^{-1}$ towards a thin convex lens of focal length $10 \,cm$ along its optical axis. When the source is $15 \,cm$ away from the lens,the image is moving at:

Similar Questions

The image distance of an object placed $10 \ cm$ in front of a thin lens of focal length $+ 5 \ cm$ is......$cm$.

The distance between an object and the screen is $100 \ cm$. $A$ lens produces an image on the screen when placed at either of the positions $40 \ cm$ apart. The power of the lens is

$A$ convex lens $T$ is used to form an image whose size is one-fourth that of the size of the object. Then the object distance is (in $f$)

The power of a thin convex lens placed in air is $+4 \ D$. The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then

$A$ hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids $L_1$ and $L_2$ having refractive indices $n_1$ and $n_2$ respectively $(n_2 > n_1 > 1)$. The lens will diverge a parallel beam of light if it is filled with

Vedclass Test Series

Exam Paper Generator

Online Exam Module