A converging lens produces a real, magnified and well defined image of

1.	A converging lens produces a real, magnified and well defined image of a small iluminatedsquare on a screen. The area of the image is A1. When the lens is moved towards the screenwithout disturbing the object and the screen, the area of the well defined image obtained onthe screen is A2. What is the side of the square object?(a) (√A1 + √A2)/2(b) [(A1+ A2)/2]^1/2(c) (A1A2)^1/2(d) (A1A2)^1/4
Answer» Answer: Option (d) $(A_1A_2)^{1/4}$ Explanation: Let the side of the SQUARE is a The positions are conjugate positions. If in the FIRST case, the length of the image is a₁ then $\frac{<klux>V</klux>}{u}=\frac{a_1}{a}$ When the LENS is moved towards the screen, v and u will be interchanged. If the image size is a₂ then $\frac{u}{v}=\frac{a_2}{a}$ From the above two equations $\frac{a_1}{a}\times\frac{a_2}{a} =1$ $\implies a_1a_2=a^2$ $\implies a=\sqrt{a_1a_2}$ But $a_1=\sqrt{A_1}$ And $a_2=\sqrt{A_2}$ Therefore, $a=\sqrt{\sqrt{A_1}\sqrt{A_2}$ $\implies a=\sqrt{(A_1A_2)^{1/2}}$ $\implies a=((A_1A_2)^{1/2})^{1/2}$ $\implies a=(A_1A_2)^{1/4}$ Thus, CORRECT option is option (d)