Ask question

Ask question

All categories

Liono4ka [1.6K]

2 years ago

14

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

ularly spaced lines. Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. If the screen is 11 m away, the spots are 4.5 cm apart, and the lines of the grating are separated by 0.5 mm, what is λ (mm)?

1 answer:

harkovskaia [24]2 years ago

3 0

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 × $10^{-2}$ m

separation L = 0.5 mm = 0.5 × $10^{-3}$ m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 × $10^{-2}$ / 11

θ = 0.234

and we know diffraction grating for max

L sinθ = mλ

here we know m = 1 so put all value and find λ

L sinθ = mλ

0.5 × $10^{-3}$ sin(0.234) = 1 λ

λ = 2042.02 × $10^{-9}$ m

λ = 2042 nm

You might be interested in

A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik

Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) $1\times10^6N/C$

B) $2\times10^{-13}N$

C) $1\times10^{-2}V$

D) Toward the inner wall

E) $3\times10^{-17}J$

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula $E=\frac{\sigma}{k\epsilon_0}$ , where $\epsilon_0=8.85\times10^{-12}C/Vm$ and in our case $\sigma=10^5C/m^2$ and, for air, $k=1.00059$ , so we have:

$E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C$

B) The K+ ion has one elemental charge excess, so its charge is $q=1.6\times10^{-19}C$ , and the force a charge experiments under an electric field E is given by F=qE, so we have:

$F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N$

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by $\Delta V=dE$ , so we have:

$\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V$

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference $\Delta V$ on a charge Q is $W=Q\Delta V$ , and is equal to the kinetic energy imparted on it, so we have:

$K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J$

5 0

2 years ago

A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the wave speed wave?

arlik [135]

Answer:

Explanation:

(1.7 m/cycle)(46 cycle/s) = 78.2 m/s

4 0

2 years ago

A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ

3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

$PE_i + KE_i = PE_f + KE_f$

$mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2$

now here we know that

$h_1 = 2.75 m$

$v_i = 0$

$v_f = 5.23 m/s$

now plug in all values in above equation

$mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2$

divide whole equation by mass "m"

$9.8*2.75 = 9.8*h + \frac{1}{2}*27.35$

$9.8*h = 13.27$

$h = 1.35 m$

so height of the ball from ground will be 1.35 m

4 0

2 years ago

How much would it cost to cover the entire land area of United States with dollar bills? To answer this question, you may find t

s2008m [1.1K]

Answer:

It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.

Explanation:

the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm

the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles

each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km

1 sq km = 1,000,000 sq meters

each sq meter = 10,000 sq cm

1 sq m = 10,000 / 100.75 = 99.25558 bills

1 sq km = 99,255,583.13 bills

9,323,964 sq km = 925,455,483,900,000 bills

8 0

2 years ago

Read 2 more answers

The following items are positioned in sequence: A source of a beam of natural light of intensity I0, three ideal polarizers A, B

statuscvo [17]

Answer:

Explanation:

Intensity of unpolarized light = I(o)

Intensity of light after first polarization by polarizer A. = I(o)/2

Angle between A and B = 120 degree.

Intensity of light after second polarization = I Cos² θ

= I(o) /2 x cos²120 = I(o) /8 .

Angle between B and C is 70 degree

Intensity of light after third polarization =

I(o)/8 x Cos² 70 = 0.1156 x I (o) /8 =

Required ratio =.01445

5 0

2 years ago