A bright object is placed on one side of a converging lens offocal length f, and

Question

A bright object is placed on one side of a converging lens offocal length f, and a white screen for viewing the image is on theopposite side. The distance d_T = d_i + d_0 between the object andthe screen is kept fixed, but the lens can be moved.

A.)Show that if d_t > 4f , there will be two positions where thelens can be placed and a sharp image can be produced on thescreen.
B.)And if d_t < 4f , no lens position where a sharrp image isformed.
C.)Also determine a formula for the distance b/w the two lensposition,and the ratio of the image sizes.
D.) and the ration of the image sizes
A bright object is placed on one side of a converging lens offocal length f, and a white screen for viewing the image is on theopposite side. The distance d_T = d_i + d_0 between the object andthe screen is kept fixed, but the lens can be moved.

A.)Show that if d_t > 4f , there will be two positions where thelens can be placed and a sharp image can be produced on thescreen.
B.)And if d_t < 4f , no lens position where a sharrp image isformed.
C.)Also determine a formula for the distance b/w the two lensposition,and the ratio of the image sizes.
D.) and the ration of the image sizes

Explanation / Answer

Given :         dT = di + do We have :       1 / do + 1 / di    = 1 /f        1 /do + 1 / ( dT -do) = 1/ f When we arrange this we get a quadratic equation fordo .              do2 - dTdo + dT f   = 0 The soution for the above equation is :             do = 1/2 [ dT ± (dT2 - 4 dT f)1/2 ] If dT   > 4f , wesee that the term inside the suareroot             dT2 - 4dTf   > 0 and          (dT2 - 4 dT f)1/2   < dT , so weget two real , positive solutions for do . (b)    If dT   < 4f , we see thatthe term inside the suareroot             dT2 - 4dTf   < 0 so there are no real solutions for do . (c)              d = do1 - do2               = 1/2 [ dT + ( dT2 - 4 dT f) 1/2 ]  - 1/2   [ dT  - (dT2 - 4 dT f)1/2 ]                 = ( dT2 - 4 dT f)1/2      di       = dT - do              = 1/2 [ dT  ± (dT2 - 4 dT f)1/2 ] So ratio of image sizes is the ratio of magnifications           m = m2 / m1               =   [ di2 / do2 ] * [do1 /   di1 ]               =  { 1/2 [ dT + (dT2 - 4 dT f)1/2 ] /   1/2   [dT  - ( dT2 - 4 dT f) 1/2 ]  }2      Hope this helps u!               di       = dT - do              = 1/2 [ dT  ± (dT2 - 4 dT f)1/2 ] So ratio of image sizes is the ratio of magnifications           m = m2 / m1               =   [ di2 / do2 ] * [do1 /   di1 ]               =  { 1/2 [ dT + (dT2 - 4 dT f)1/2 ] /   1/2   [dT  - ( dT2 - 4 dT f) 1/2 ]  }2                   = 1/2 [ dT  ± (dT2 - 4 dT f)1/2 ] So ratio of image sizes is the ratio of magnifications           m = m2 / m1               =   [ di2 / do2 ] * [do1 /   di1 ]               =  { 1/2 [ dT + (dT2 - 4 dT f)1/2 ] /   1/2   [dT  - ( dT2 - 4 dT f) 1/2 ]  }2      Hope this helps u!         

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