The distance between the object and the scree | vedclass

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

Question

(B) The displacement method formula for the focal length $f$ of a lens is given by $f = \frac{D^2 - x^2}{4D}$,where $D$ is the distance between the ob

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The displacement method formula for the focal length $f$ of a lens is given by $f = \frac{D^2 - x^2}{4D}$,where $D$ is the distance between the object and the screen,and $x$ is the distance between the two positions of the lens.Given $D = 100 \, cm$ and $x = 40 \, cm$.Substituting these values: $f = \frac{100^2 - 40^2}{4 	imes 100} = \frac{10000 - 1600}{400} = \frac{8400}{400} = 21 \, cm$.The power $P$ of the lens in Diopters $(D)$ is given by $P = \frac{100}{f(cm)}$.$P = \frac{100}{21} \approx 4.76 \, D$.Rounding to the nearest integer,we get $P \approx 5 \, D$.

The distance between the object and the screen is $100 \, cm$. $A$ lens produces an image on the screen when it is placed at either of two positions $40 \, cm$ apart. The power of the lens is (approximately): (in $, D$)

Similar Questions

What is a lens?

$A$ concave lens of focal length $f$ produces an image $(1/x)$ of the size of the object. The distance of the object from the lens is:

Two convex lenses of focal lengths $f_1$ and $f_2$ form images with magnification $m_1$ and $m_2$, when used individually for an object kept at the same distance from the lenses. Then, $f_1 / f_2$ is

If the focal lengths of a lens for red and violet light rays are $f_{R}$ and $f_{V}$ respectively,then which of the following is the true relationship?

An optician prescribes a corrective lens of power $+2.0 \,D$. The focal length of the required convex lens is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module