Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

A luminous object and a screen are at a fixed distance ’D’ apart. A converging lens for focal length 'f' is placed between the object and screen. A real image of the object is formed on the screen for two less positions separated by a distance’d’. The ratio of two image sizes for two positions of the lens is:

Option: 1

\left ( \frac{{D-d}}{D+d} \right )^{2}


Option: 2

\left ( \frac{D+d}{D-d} \right )^{2}


Option: 3

\left ( \frac{D-2d}{D+2d} \right )^{2}


Option: 4

\left ( \frac{D+2d}{D-2d} \right )^{2}


Answers (1)

best_answer

As we learn

Magnification at two position of lens -

m_{1}= \frac{D\, +d}{D-d}

m_{2}= \frac{D\, -d}{D+d}
 

- wherein

m_{1}=Mangnification at position 1 of lens.

m_{2}=Mangnification at position 2

 

 M_{1}=\frac{D+d}{D-d}                    M_{2}=\frac{D-d}{D+d}

\frac{M_{2}}{M_{1}}=\frac{\left ( D-d \right )*\left ( D-d \right )}{\left ( D+d \right )*\left ( D+d \right )}=\left ( \frac{D-d}{D+d} \right )^{2}

 

Posted by

Nehul

View full answer

Similar Questions

Latest Question