A converging lens of focal length 20.0 cm forms images of an object situated at

Question

A converging lens of focal length 20.0 cm forms images of an object situated at various distances. (a) If the object is placed 40.0 cm from the lens, locate the image, state whether its real or virtual, and find its magnification. (b) Repeat the problem when the object is at 20.0 cm and (c) again when the object is 10.0 cm from the lens. (d) Follow through with ray tracing. Repeat Problem but now for a diverging lens of focal length 10.0 cm. Similarly, draw a ray trace to show the effect of a diverging lens.

Explanation / Answer

v = (f*u)/(u-f) = 20*40/(40-20) = 40cm...real and inverted..
m = -v/u = 4-0/40 = -1 ...so same size as image...

b) v = (f*u)/(u-f) = 20*20/(20-20) = infinite...image not forms...

C) v = (f*u)/(u-f) = 10*20/(10-20) = -20 cm cm...virtual and erected..
m = v/u = 20/10 = 2..so image is enlarged..
d) where we keep the object the final image forms between focal point and optic center
for 40 cm ...v = -(f*u)/(f+u) = -10*40/(10+40) = -8 cm
for 20 cm.....v = -(f*u)/(f+u) = -10*20/(10+20) = -6.67 cm
for 10 cm...v = -(f*u)/(f+u) = -10*10/(10+10) = -5 cm

the formed image is virtual,erect and dimished

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.