Two point light sources are 24 , cm apart. Wh | vedclass

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

Question

(A) Let the distance of the first source $S_1$ from the lens be $x$. Then the distance of the second source $S_2$ from the lens is $(24 - x)$.For the

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) Let the distance of the first source $S_1$ from the lens be $x$. Then the distance of the second source $S_2$ from the lens is $(24 - x)$.For the images to be formed at the same place,one source must form a virtual image on the same side,and the other must form a real image on the opposite side.Let the virtual image of $S_1$ be at distance $y$ from the lens on the left side. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{9} = \frac{1}{-y} - \frac{1}{-x} \Rightarrow \frac{1}{y} = \frac{1}{x} - \frac{1}{9}$ .....$(i)$Let the real image of $S_2$ be at distance $y$ from the lens on the left side. Using the lens formula:$\frac{1}{9} = \frac{1}{-y} - \frac{1}{-(24 - x)} \Rightarrow \frac{1}{y} = \frac{1}{24 - x} - \frac{1}{9}$ .....$(ii)$Equating $(i)$ and $(ii)$:$\frac{1}{x} - \frac{1}{9} = \frac{1}{24 - x} - \frac{1}{9} \Rightarrow \frac{1}{x} = \frac{1}{24 - x} \Rightarrow x = 24 - x \Rightarrow 2x = 24 \Rightarrow x = 12 \, cm$.Wait,the condition for images to coincide is that the virtual image of one and real image of other coincide. Let $S_1$ be at $u_1 = -x$ and $S_2$ be at $u_2 = -(24-x)$.For $S_1$,$v_1 = -y$. $\frac{1}{9} = \frac{1}{-y} - \frac{1}{-x} \Rightarrow \frac{1}{y} = \frac{1}{x} - \frac{1}{9}$.For $S_2$,$v_2 = +y$. $\frac{1}{9} = \frac{1}{y} - \frac{1}{-(24-x)} \Rightarrow \frac{1}{y} = \frac{1}{9} - \frac{1}{24-x}$.Equating: $\frac{1}{x} - \frac{1}{9} = \frac{1}{9} - \frac{1}{24-x} \Rightarrow \frac{1}{x} + \frac{1}{24-x} = \frac{2}{9}$.$\frac{24-x+x}{x(24-x)} = \frac{2}{9} \Rightarrow \frac{24}{24x - x^2} = \frac{2}{9} \Rightarrow 108 = 24x - x^2 \Rightarrow x^2 - 24x + 108 = 0$.Solving the quadratic: $(x - 6)(x - 18) = 0$. Thus $x = 6 \, cm$ or $18 \, cm$.

Two point light sources are $24 \, cm$ apart. Where should a convex lens of focal length $9 \, cm$ be placed between them from one source so that the images of both the sources are formed at the same place (in $, cm$)?

Similar Questions

The minimum distance between an object and its real image formed by a convex lens is

$A$ double convex lens has power $P$ and same radii of curvature $R$ of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power $1.5P$ is ......$R$.

$A$ 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

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it,then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

When a biconvex lens of glass having refractive index $1.47$ is dipped in a liquid,it acts as a plane sheet of glass. This implies that the liquid must have a refractive index:

Vedclass Test Series

Exam Paper Generator

Online Exam Module