A bright object is placed on one side of a converging lens offocal length f, and
ID: 1745703 • Letter: A
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!Related Questions
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