An object approaches a convergent lens from t | vedclass

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

Question

(C) From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Differentiating with respect to time $t$,we get $0 = -\frac{1}{v^2} \frac{dv}{dt}

Vedclass · Accepted Answer

moves away from the lens with a non-uniform acceleration

Vedclass · Answer

(C) From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Differentiating with respect to time $t$,we get $0 = -\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt}$.Rearranging,we find the velocity of the image $v_i = \frac{dv}{dt} = \left(\frac{v^2}{u^2}ight) \frac{du}{dt} = m^2 v_o$,where $m = \frac{v}{u}$ is the magnification and $v_o$ is the object speed.Since $m = \frac{f}{f+u}$,the magnification $m$ changes as the object moves towards the focus $(u 	o -f)$.Differentiating the velocity $v_i$ with respect to time to find acceleration $a_i = \frac{dv_i}{dt} = \frac{d}{dt}(m^2 v_o) = 2m v_o \frac{dm}{dt}$.As the object approaches the focus,$m$ changes non-linearly,making $\frac{dm}{dt}$ variable. Thus,the image moves away from the lens with a non-uniform acceleration.

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \ m/s$ and stops at the focus. The image:

Similar Questions

The focal length of a convex lens of refractive index $1.5$ is $2 \ cm$. What will be the focal length of the lens when it is immersed in a liquid of refractive index $1.25$?

If the focal length of a double convex lens for red light is $f_R$,what is its focal length for violet light?

The focal length of a convex lens depends upon

$A$ thin convex lens of focal length $f$ made of crown glass is immersed in a liquid of refractive index $\mu_l$ $(\mu_l > \mu_c)$, where $\mu_c$ is the refractive index of the crown glass. The convex lens now acts as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module