A lens having focal length f and aperture of | vedclass

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

Question

(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering the central pa

Vedclass · Accepted Answer

$f$ and $3I/4$

Vedclass · Answer

(D) The focal length $f$ of a lens depends on the refractive index of the material and the radii of curvature of its surfaces. Covering the central part of the lens does not change these parameters,so the focal length remains $f' = f$.Intensity $I$ is proportional to the area of the aperture $A$. The initial area is $A = \pi (d/2)^2 = \pi d^2 / 4$.The area of the covered central part is $A_{covered} = \pi (d/4)^2 = \pi d^2 / 16$.The remaining area is $A' = A - A_{covered} = \frac{\pi d^2}{4} - \frac{\pi d^2}{16} = \frac{3\pi d^2}{16}$.Since $I \propto A$,the new intensity $I'$ is given by $I' = I 	imes (A'/A) = I 	imes (3/4) = 3I/4$.

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. If the central region of the lens with diameter $\frac{d}{2}$ is covered by black paper,what will be the new focal length and intensity of the image,respectively?

Similar Questions

$A$ neutrino is a particle,which is

$f(x) = (20 - x^4)^{1/4}$ for $0 < x < \sqrt{5}$,then $f(f(1/2))$ is equal to

The Bhopal gas tragedy of $1984$ was caused by:

The silicates which contain $Si_2O_7^{6-}$ units are called

Which one of the following statements regarding helium is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module