Ask question

Ask question

All categories

3 years ago

10

A transparent oil with index of refraction 1.15 spills on the surface of water (index of refraction 1.33), producing a maximum o

f reflection with normally incident violet light (wavelength 400 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick.

1 answer:

Zolol [24]3 years ago

5 0

Answer:

The thickness of the oil slick. t = 173.91 nm

Explanation:

Oil film thickness t is given by the equation

$t = \frac{\lambda}{2n}$

where λ = wavelength of incident light in air = 400 nm

and n = index of refraction of oil

therefore,

$t =\frac{400}{2\times 1.15}\\t= 173.91 nm$

The thickness of the oil slick. t = 173.91 nm

You might be interested in

What provides the force that causes magma to erupt to the surface

olga2289 [7]

Geothermal energy provides force that causes magma to erupt to the surface

3 0

3 years ago

Air passing over an airplane's wing travels ____________________, and therefore exerts ____________________ pressure than air tr

Aleksandr [31]

<span>c. faster, less

is the answer
</span>

6 0

3 years ago

Read 2 more answers

Use the dot product to find the magnitude of u if u = 6i - 3j

LuckyWell [14K]

Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | = $\sqrt{6 ^{2}+(-3) ^{2} } = \sqrt{36+9}= \\ \sqrt{45}= \sqrt{9*5}=3 \sqrt{5}$
Answer:
The magnitude of vector u is 3√5.

3 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t

pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

$x = v_0 t +\frac{1}{2}at^2$

$48.1 = 0*t + \frac{1}{2} (9.8)t^2$

$t = 3.13s$

The second object is thrown downward at one second later and it meets the first object at the water is

$t' = 3.13 -1.48$

$t' = 1.65s$

The distance of the object will travel due to free fall acceleration is

$x = v_0 t+\frac{1}{2} at^2$

$x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2$

$x = 13.34m$

The distance of the object will travel due to its initial velocity is

$v_0 = \frac{d_0}{t}$

$d_0 = v_0 t$

$48.1-13.34 = v_0 (1.65)$

$v_0 = 21.06m/s$

Therefore the initial speed of the second object is 21.06m/s

8 0

4 years ago