Question

A symmetric double convex lens in cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4 D, the power a cut-lens will be
(a) 2 D
(b) 3 D
(c) 4 D
(d) 5 D.

Answer 1

power a cut-lens will be  $2D$

Explanation:

When we cut a doble convex lens through perpendicular to the principal axis of lens, every part of lens behave as plano- convex lens.

doble convex lens  redii are $R_1 = r$, $R_2 = -r$

Refractive index of double convex lens $= \mu$

Focal length of double convex lens $\frac{1}{f}$$= (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$

⇒   $\frac{1}{f_d} = (\mu-1) \frac{2}{r}$

⇒   $f_d = \frac{r}{2(\mu-1)}$

plano-convex lens redii are  $R_1 = r$, $R_2 = \infty$

refractive index will be same because it is the property of medium

$\frac{1}{f_p}=(\mu - 1)(\frac{1}{r} - \frac{1}{\infty})$

⇒ $f_p = \frac{r}{(\mu - 1)}$

On Comparing the values of $f_d$ and $f_p$ we get, $f_p = 2 f_d$

As we know that, power of lens $P = \frac{1}{\textrm{focal length}}$

$\frac{P_p}{P_d} = \frac{f_d}{f_p}$

power of plano-convex lens $P_p = \frac{4}{2} = 2 Diopter$

Answer 2

Power of cut lens will be 2D

Explanation:

• Biconvex lens is cut perpendicularly to the principal axis, it will become a plano- convex lens.

• Biconvex Lens has Focal Length -

$\frac {1}{f} = (n - 1)(\frac {1}{R_1} - \frac {1}{R_2})$

$\frac {1}{f} = (n - 1)\frac {2}{R} (As R_1 = R & R_2 = -R)$

$\frac {1}{f} = \frac {R}{2}(n - 1)$ """(i)

• For piano convex lens

$\frac {1}{f_1} =(n - 1)(\frac {1}{R} - \frac {1}{\infty })$

$f_1 = \frac {R}{n - 1}$ """"" (ii)

• Comparing eqns (i) & (ii) the focal length gets double

• As power of lens $P \infty = \frac {1}{focal\ length}$

Hence, Power will become half.

New Power = $\frac {4}{2}$

= 2D