Question

Note: We suggest you use ray diagrams to qualitatively understand these questions. A candle 6.10 cm high is placed in front of a thin converging lens of focal length 33.0 cm. What is the image distance i when the object is placed 93.0 cm in front of the same lens?

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

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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'.)

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The object is now moved to 41.0 cm in front of the lens, what is the new image distance i?

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What is the new size of the image?

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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'.)

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The object is now moved to 21.0 cm in front of the lens, what is the new image distance i?

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What is the new size of the image?

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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'.)

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The object is now moved to 6.5 cm in front of the lens, what is the new image distance i?

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What is the new size of the image?

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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'.)

Answer 1

For the first scenario (p = 93)

1/f = 1/p + 1/q

1/33 = 1/93 + 1/q

q = 51.2 cm

Image size...

h'/h = -q/p

h'/6.1 = -(51.2)/93

h' = -3.36 cm (Negative means image is inverted but 3.36 cm tall)

Since q is positive, the image is real

Thus, the answers to the next part are RISB

For the second scenario (p = 41)

1/f = 1/p + 1/q

1/33 = 1/41 + 1/q

q = 169 cm

Image size...

h'/h = -q/p

h'/6.1 = -(169)/41

h' = -25.2 cm (Negative means image is inverted but 25.2 cm tall)

Since q is positive, the image is real

Thus, the answers to the next part are RILB

For the third scenario (p = 21)

1/f = 1/p + 1/q

1/33 = 1/21 + 1/q

q = -57.8 cm

Image size...

h'/h = -q/p

h'/6.1 = -(-57.8)/21

h' = 16.8 cm

Since q is negative, the image is virtual

Thus, the answers to the next part are VULF

For the fourth scenario (p = 6.5)

1/f = 1/p + 1/q

1/33 = 1/6.5 + 1/q

q = -8.09 cm

Image size...

h'/h = -q/p

h'/6.1 = -(8.09)/6.5

h' = 7.60 cm

Since q is negative, the image is virtual

Thus, the answers to the next part are VULF