A concave mirror has a radius of curvature of 1 . 57 m. Calculate the position o

Question

A concave mirror has a radius of curvature of 1.57 m.

Calculate the position of the image pro- duced when an object is placed 2.08 m from the mirror.

Answer in units of m

Calculate the position of the image when an object is placed 0.505 m from the mirror.

Answer in units of m

The top of a swimming pool is at ground level. If the pool is 2.4 m deep, how far below ground level does the bottom of the pool ap- pear to be located when the pool is completely

filled with water? Answer in units of m

How far below ground level does the bottom of the pool appear to be located when the pool is filled halfway with water?

Answer in units of m

Explanation / Answer

1)

a).

The object is between F and O therefore the image is virtual. Values concerning the image will thus be negative.
1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = (u - f)/f*u
v = f*u/(u - f)
v = 1.57m*2.08m/(2.08m - 1.57m)
v = 6.40m

where
F = focus
O = physical centre of mirror
u = object distance from O
v = image distance from O
f = focal length (radius of curvature)

b).

now u = 0.505m

so again putting values we get

v = f*u/(u-f)

v = 1.57*0.505/(0.505-1.57)

v = -0.744m

2).

Lets assume the refractive index is 1.33 for water.

a)

Just use the formula n = real depth / apparent depth.

1.33 = 2.4/apparent depth

Apparent depth = 1.80m

Therefore, the floor of the pool is seen to be 1.80m below ground level.

b)

Since the water is only half way up the pool, refraction will occur at the water surface half of the pool. Now, the real depth is only 1.2m.

1.33 = 1.2/apparent depth

apparent depth = 0.90m

Therefore, the total depth is 1.80 + 0.90 = 2.7m below ground level.

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