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3 years ago

Help I need help with this

Physics

1 answer:

igomit [66]3 years ago

3 0

Answer:

8 meters per second

Explanation:

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The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th

pickupchik [31]

Answer:

The focal length of the lens in water is $f_{water} = 262.68 cm$

The focal length of the mirror in water is $f =79.0cm$

Explanation:

From the question we are told that

The index of refraction of the lens material = $n_2$

The index of refraction of the medium surrounding the lens = $n_1$

The lens maker's formula is mathematically represented as

$\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

Where $f$ is the focal length

$n$ is the index of refraction

$R_1 and R_2$ are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air we have

$\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

When immersed in liquid the formula becomes

$\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]$

The ratio of the focal length of the the two medium is mathematically evaluated as

$\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }$

From the question

$f_{air }$ = 79.0 cm

$n_2 = 1.55$

and the refractive index of water(material surrounding the lens) has a constant value of $n_1 = 1.33$

$\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}$

$f_{water} = 262.68 cm$

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

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If a bowling ball rolls 12 meters in 2.2 seconds what is the speed ?

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A weightlifter deadlifts a 300 kg weight. If the weightlifter has a mass of 100 kg, what is the force acting on his legs?

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