For a convex lens,the distance of the object | vedclass

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

Question

(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $v$ is the image distance,$u$ is the object distance,and $f$ is the f

Vedclass · Accepted Answer

Hyperbola

Vedclass · Answer

(D) The lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $v$ is the image distance,$u$ is the object distance,and $f$ is the focal length of the lens.Let $x = u$ and $y = v$. Then the equation becomes $\frac{1}{y} - \frac{1}{x} = \frac{1}{f}$.Rearranging the terms,we get $\frac{1}{y} = \frac{1}{x} + \frac{1}{f} = \frac{x+f}{xf}$.Thus,$y = \frac{xf}{x+f}$.This equation represents a rectangular hyperbola shifted from the origin. Therefore,the graph obtained by plotting the image distance against the object distance is a hyperbola.

For a convex lens,the distance of the object is taken on the $X$-axis and the distance of the image is taken on the $Y$-axis. The nature of the graph so obtained is:

Similar Questions

$A$ convex lens has a focal length of $0.3 \ m$ and a refractive index of $3/2$. Find the focal length of the lens in $m$ when it is immersed in water,which has a refractive index of $4/3$.

The focal length of a thin biconvex lens is $20 \ cm$. When an object is moved from a distance of $25 \ cm$ in front of it to $50 \ cm$,the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

The optical axis of a thin equiconvex lens is the $x$-axis. The coordinates of the object and its image are $(-40 \, cm, 1 \, cm)$ and $(50 \, cm, -2 \, cm)$ respectively. Find the position of the lens.

Define power of a lens,obtain its equation and write its $SI$ unit.

$A$ biconvex lens has radii of curvature,$20 \ cm$ each. If the refractive index of the material of the lens is $1.5$,the power of the lens is .... $D$

Vedclass Test Series

Exam Paper Generator

Online Exam Module