1) At a height of ten meters above the surface of a freshwater lake, a sound pul

Question

1) At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the lake returns to the point of origin 0.151 s later. The air and water temperatures are 20 oC. How deep is the lake?

2)When a diverging lens is held 13.4 cm above a line of print, as in Figure below, the image is 5.7 cm beneath the lens. (a) Is the image real or virtual? (b) What is the focal length of the lens?

3)Two identical diverging lenses are separated by 23 cm. The focal length of each lens is -13 cm. An object is located 5.1 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

4)A diverging lens (f =

Explanation / Answer

1)Time = distance /speed.
Speed of sound in air = 343.2m/s at 20?C
Speed of sound in water = 1484.3 m/s at 20?C
The time for the sound to travel 2*10 m is 20 / 343.2
Time to travel d m in water =2d/1484.3
2d/1484.3 + 20 / 343.2 = 0.151s (Given)

d = 68.1 m

2)a) virtual

b)1/f = 1/u + 1/v
Since this is a virtual image,
1/f = 1/(-13.4) + 1/5.7
1/f = 0.564
f = 1.77cm (ans)
d = 68.1 m

3) e the thin lens formula to find the image formed by first lens;

1/5.1 + 1/q = -1/13

q = -3.70, the negative means the image is on the left of the first lens and its a virtual image.

Now use this image as your object for the second lens. It is 3.70 + 23 = 26.70 away from the 2nd lens

1/26.7 + 1/q = -1/16

q = - 10.10cm

This is the image formed by the 2nd lens. it is 10.10 cm to the left of the 2nd lens and is also a virtual image.

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