**SELECT BEST ANSWER** 1. A 3.0cm tall object is located 25.0cm from a convergin
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Question
**SELECT BEST ANSWER**
1. A 3.0cm tall object is located 25.0cm from a converging lens that has a focal length of 10.0cm. Where is the image located and what type of image is it?
(a) On the same side as the object, 16.7 cm from the lens. Real and Upright
(b) On the opposite side as the object, 16.7 cm from the lens. Virtual and Upright
(c) On the same side as the object, 16.7 cm from the lens. Real and Inverted
(d) On the opposite side as the object, 16.7 cm from the lens. Virtual and Inverted
(e) On the opposite side as the object, 16.7 cm from the lens. Real and Inverted
2. A 3.0cm tall object is located 5.0cm from a converging lens that has a focal length of 10.0cm. Where is the image located and what type of image is it?
(a) On the same side as the object, 10.0 cm from the lens. Real and Upright
(b)On the same side as the object, 10.0 cm from the lens. Virtual and Upright
(c) On the same side as the object, 10.0 cm from the lens. Real and Inverted
(d) On the opposite side as the object, 10.0 cm from the lens. Virtual and Inverted
(e) On the opposite side as the object, 10.0 cm from the lens. Virtual and Upright
3. A 3.0cm tall object is located 5.0cm from a diverging lens that has a focal length of 10.0cm. Where is the image located and what type of image is it?
(a) On the same side as the object, 5.00 cm from the lens. Real and Upright
(b) On the same side as the object, 5.00 cm from the lens. Virtual and Upright
(c) On the same side as the object, 5.00 cm from the lens. Real and Inverted
(d) On the opposite side as the object, 5.00 cm from the lens. Virtual and Inverted
(e) On the opposite side as the object, 5.00 cm from the lens. Virtual and Upright
Explanation / Answer
object distance = U = 25 cm
focal length = F = 10 cm
image distance = V = ?
by the lens maker's formula
we have
1/F = 1/U + 1/V
1/V = 1/F - 1/U
= 1/10 - 1/25
V =+ 16.7 cm
On the opposite side as the object, 16.7 cm from the lens. Real and Inverted
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