Please show work, thanks :) A diverging lens has a focal length of -10.00 cm. (a

Question

Please show work, thanks :)

A diverging lens has a focal length of -10.00 cm.

(a) What are the image distances q for objects placed at the distances p from the lens which are tabulated below? In each case, calculate the magnification m, describe the image as real or virtual, upright or inverted, and enlarged or diminished in size.

(b) If the object is 4.00 cm high, what is the height of the image for the object distances of 5.00 cm and 20.0 cm?

object distance (cm) 5.00 10.00 14.0 16.0 20.0 image distance (cm) magnification real or virtual? ---Select--- real virtual ---Select--- real virtual ---Select--- real virtual ---Select--- real virtual ---Select--- real virtual upright or inverted? ---Select--- upright inverted ---Select--- upright inverted ---Select--- upright inverted ---Select--- upright inverted ---Select--- upright inverted enlarged or diminished? ---Select--- enlarged diminished ---Select--- enlarged diminished ---Select--- enlarged diminished ---Select--- enlarged diminished ---Select--- enlarged diminished

Explanation / Answer

We know that the diverging lens can produce always

the image is VIRTUAL , UPRIGHT , DIMINISHED IMAGES

Given focal length is f = -10 cm, object distance p , image distance q, magnification is m = -q/p

the relation between p,q,f is , 1/f = 1/p +1/q==> 1/q = 1/f-1/p

iamge distance

for p=5 cm, 1/q = -1/10 -1/5,q = -3.33 cm ;

for p=10cm, 1/q = -1/10 -1/10 = -5 cm;

for p=14 cm, 1/q = -1/10 -1/14 = -5.833 cm;   

for p = 16 cm, 1/q = -1/10 -1/16 = -6.154 cm;

for p = 20 cm, 1/q = -1/10 -1/20 = -6.66 cm;

Magnification m= -q/p = -(-3.33)/5 = 0.666 ;m = -(-5)/10= 0.5;m = -(-5.833)/14= 0.41664; m= -(-6.154)/16= 0.384625;m = -(-6.66)/20 = 0.333

====>

In all the cases the image is virtual image and upright image

In all the cases the image is diminished..

b)

p = 5c, , y = 4 cm and f = -10 cm

1/q = 1/f - 1/p = -1/10 - 1/5==> q = -3.33 cm

for p=10 cm , 1/q = -1/10 -1/10 = -5 cm

now height of the image is m= -q/p= y'/y ==>y' = -q*y/p

for p=5 cm , y' = -(-3.33)*4/5 cm = 2.664 cm

for p=10 cm , y' = -(-5)*4/10 cm = 2 cm

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