A 5 mm high pin is placed at a distance of 15 cm from a convex lens of focal length 10 cm. A second lens of focal length 5 cm is placed 40 cm from the first lens and 55 cm from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.
Answers
Given:
height of pin=ho=5mm
u= -30cm
f=10cm
Distance between two lens=40cm
second lens focal length=5cm
Distance between second lens and pin=55cm
a)
Image formed by first lens:
u=-15cm
f= 10cm
by lens formula:
1/f=1/v-1/u
1/v=1/f+1/u=1/10-1/15 = 3- 2/30=1/30
v=30cm
This image will become object for second lens
object distance=u= -(40-30)=-10cm
f1=5cm
lens formula:
1/f=1/v-1/u
1/v=1/f+1/u
1/v=1/5-1/10
1/v=1/10
v=10cm
Final position of image is 10cm right from second lens
b)magnification by first lens:
m=v/u=hi/ho
hi=-5x30/15
hi==10mm
m magnification by second lens
hfinal/hi=v/u
10/-10=hfinal/-10
hfinal=10mm
Thus image is erect
c) size of final image is 10mm
Image formed by first lens:
u=-15cm
f= 10cm
by lens formula:
1/f=1/v-1/u
1/v=1/f+1/u=1/10-1/15 = 3- 2/30=1/30
v=30cm
This image will become object for second lens
object distance=u= -(40-30)=-10cm
f1=5cm
lens formula:
1/f=1/v-1/u
1/v=1/f+1/u
1/v=1/5-1/10
1/v=1/10
v=10cm
Final position of image is 10cm right from second lens
b)magnification by first lens:
m=v/u=hi/ho
hi=-5x30/15
hi=10mm
m magnification by second lens
hfinal/hi=v/u
10/-10=hfinal/-10
hfinal=10mm
Thus image is erect
c) size of final image is 10mm