A lens is placed between a source of light an | vedclass

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

Question

(A) Let the area of the source be $A_0$. For the two positions of the lens,the magnification $m_1$ and $m_2$ are given by $m_1^2 = \frac{A_1}{A_0}$ an

Vedclass · Accepted Answer

$\sqrt{A_1 A_2}$

Vedclass · Answer

(A) Let the area of the source be $A_0$. For the two positions of the lens,the magnification $m_1$ and $m_2$ are given by $m_1^2 = \frac{A_1}{A_0}$ and $m_2^2 = \frac{A_2}{A_0}$.From the property of conjugate positions of a lens,the product of the magnifications is $m_1 	imes m_2 = 1$.Therefore,$(m_1 	imes m_2)^2 = 1^2 = 1$.Substituting the expressions for $m_1^2$ and $m_2^2$,we get $\frac{A_1}{A_0} 	imes \frac{A_2}{A_0} = 1$.This simplifies to $A_0^2 = A_1 A_2$.Thus,the area of the source is $A_0 = \sqrt{A_1 A_2}$.

$A$ 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

Similar Questions

$A$ lens of focal power $0.5 D$ is

The magnification of an object placed in front of a convex lens of focal length $20 \, cm$ is $+2$. To obtain a magnification of $-2$,the object has to be moved a distance equal to.....$cm$

$A$ bi-convex lens made of glass (refractive index $1.5$) is put in a liquid of refractive index $1.7$. Its focal length will

$A$ convex lens of focal length $0.15 \,m$ is made of a material of refractive index $\frac{3}{2}$. When it is placed in a liquid, its focal length is increased by $0.225 \,m$. The refractive index of the liquid is

$A$ thin plano-convex lens of focal length $73.5 \ cm$ has a circular aperture of diameter $8.4 \ cm$. If the refractive index of the material of the lens is $\frac{5}{3}$, then the thickness of the lens is nearly

Vedclass Test Series

Exam Paper Generator

Online Exam Module