Question

NOTE: PLEASE ONLY ANSWER PARTS C AND E-I. THANKS!

A candle 7.60cm high is placed in front of a thin converging lens of focal length 38.5cm.

a. What is the image distance i when the object is placed 108.5 cm in front of the same lens?

b. What is the size of the image? (Note: an inverted image will have a `negative' size.)

c. Is the image real(R) or virtual(V)?; upright(U) or inverted(I)?; larger(L) or smaller(S) or unchanged(UC)?; In front of the lens(F) or behind the lens(B)? Answer these questions in the order that they are posed. (for example, if the image is real, inverted, larger, and behind the lens then enter `RILB'.)

d. The object is now moved to 54.5 cm in front of the lens. What is the new image distance i?

e. What is the new size of the image? ( Note, an inverted image will have a `negative' size.)

f. Is the new image real(R) or virtual(V)?; upright(U) or inverted(I)?; larger(L) or smaller(S) or unchanged(UC)?; In front of the lens(F) or behind the lens(B)? Answer these questions in the order that they are posed. (for example, if the image is real, inverted, larger, and behind the lens then enter `RILB'.)

g. The object is now moved to 24.5 cm in front of the lens. What is the new image distance i?

h. What is the new size of the image? ( Note, an inverted image will have a `negative' size.)

i. Is the new image real(R) or virtual(V)?; upright(U) or inverted(I)?; larger(L) or smaller(S) or unchanged(UC)?; In front of the lens(F) or behind the lens(B)? Answer these questions in the order that they are posed. (for example, if the image is real, inverted, larger, and behind the lens then enter `RILB'.)

NOTE: PLEASE ONLY ANSWER PARTS C AND E-I. THANKS!

Answer 1

solution

GIVEN DATAS:

h0= 7.6 cm ,f=38.5 ,d0=108.5

NOW, USING FORMULA:

c )1/f=1/d0+1/di

=>1/38.5=1/108.5+1/di

=>1/di=0.0167

=>di= 59.88 cm

E) hi/h0= -di/d0

=>hi/7.60= -59.88/108.5

=>hi= -4.19