An unsymmetrical double convex thin lens form | vedclass

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

Question

(B) According to the lens maker's formula,$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.From the lens formula,$\frac{1}{v} - \f

Vedclass · Accepted Answer

No

Vedclass · Answer

(B) According to the lens maker's formula,$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.From the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.When the lens is reversed,the new radii of curvature become $R_1' = -R_2$ and $R_2' = -R_1$. Substituting these into the lens maker's formula:$\frac{1}{f'} = (n - 1) \left( \frac{1}{-R_2} - \frac{1}{-R_1} \right) = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{f}$.Since the focal length $f$ remains unchanged and the object distance $u$ is constant,the image distance $v$ remains unchanged according to the lens formula $\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$.Therefore,the position of the image will not change.

An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

Similar Questions

$A$ thin glass (refractive index $1.5$) lens has an optical power of $-5\,D$ in air. Its optical power in a liquid medium with a refractive index of $1.6$ will be: (in $,D$)

When a plane wavefront is incident on a convex lens,the emerging wavefront becomes . . . . . . .

$A$ thin lens made of glass of refractive index $1.5$ has a front surface $+ 11 D$ power and back surface $-6 D.$ If this lens is submerged in a liquid of refractive index $1.6,$ the resulting power of the lens is

Write the lens maker's equation for a thin lens.

The unit of focal power of a lens is

Vedclass Test Series

Exam Paper Generator

Online Exam Module