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(picture attached) use grass method

4. A block of unknown substance is submerged in water. A light ray in the water strikes the angle of 16°, what is the index of refraction of the unknown substance? (nwater = 1.33) substance at an angle of 45° from the normal. If the refracted ray in the substance is at an 5. If a light ray passes from a substance with low index of refraction to another substance with high index of refraction, will the ray bend away from or closer to the normal? Why? 6. What is the angle of refraction? n₁=1 24° n₂ = 1.8​

(picture attached) use grass method 4. A block of unknown substance is submerged in-example-1
User Smarteist
by
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1 Answer

4 votes

Answer:

Step-by-step explanation:

Step 1

1 of 2

\textbf{Given}Given

n_{w}=1.33n

w

=1.33

$\theta_w=31\text{ ^\circ

}$

$\theta_b=27\text{ ^\circ

}$

\textbf{Approach}Approach

In this problem we are going to use Snell's law.

\textbf{Solution}Solution

The definition of Snell's law of refraction is

\begin{align} {n_1\cdot\sin \theta_1}={n_2\cdot \sin \theta_2} \end{align}

n

1

⋅sinθ

1

=n

2

⋅sinθ

2

where indexes 11 and 22 represent two different mediums. Since the motion is in the water we write

\begin{align} &{n_{w}\cdot\sin\theta_{w}}={n_{b}\cdot\sin\theta_{b}} \\ &{n_b}={n_{w}\cdot \frac{\sin\theta_{w}}{ \sin \theta_{b}} } \\ &{n_b}={1.33\cdot \frac{\sin31^\circ}{ \sin27^\circ} } \\ &\boxed{{n_b}=1.5} \end{align}

n

w

⋅sinθ

w

=n

b

⋅sinθ

b

n

b

=n

w

sinθ

b

sinθ

w

n

b

=1.33⋅

sin27

sin31

n

b

=1.5

User Suraj Kochale
by
4.6k points