You visited us 1 times! Enjoying our articles? Unlock Full Access!

Cutting of Thin Lenses

A lens is pla...

Question

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

\frac{A_{1} - A_{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{A_{1}} + \frac{1}{A_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{A_{1} + A_{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{A_{1} + A_{2}}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\sqrt{A_{1} + A_{2}}$
$\begin{matrix} B y d i s p l a c e m e n t m e t h o d, s i z e o f o b j e c t (0) = \sqrt{l_{1} \times l_{2}} T h e r e f o r e a r e a o f s o u r c e o f l i g h t (A) = \sqrt{(A_{1} + A_{2})} \end{matrix}$
Hence, Option $C$ is correct.

Suggest Corrections

Similar questions

a_{1}

a_{2}

a_{3}

a_{n}

A . P .

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{(n - 1)}{\sqrt{a_{n}} + \sqrt{a_{1}}}

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

\frac{a_{1}}{a_{1} + 1 -} \frac{a_{2}}{a_{2} + 1 -} \frac{a_{3}}{a_{3} + 1 -} \dots, a_{1}, a_{2}, a_{3}, \dots

A_{1}, A_{2},

A_{3}

R^{2}

A_{1} = {(x, y) : x \geq 0, y \geq 0, 2 x + 2 y - x^{2} - y^{2} > 1 > x + y}, A_{2} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{2} + y^{2}}, A_{3} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{3} + y^{3}} .

| A_{1} |, | A_{2} |,

| A_{3} |

A_{1}, A_{2}

A_{3}

\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{4}} + . . . . . + \frac{a_{n - 1}}{a_{n}} + \frac{a_{n}}{a_{1}} > n

a_{1}, a_{2}, . . ., a_{n}

Join BYJU'S Learning Program