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Pavel [41]
2 years ago
13

In Young’s two slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.44˚. What is the r

atio of the slit separation, d, to the wavelength of the light, λ
Physics
1 answer:
melomori [17]2 years ago
8 0

Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

   d sin θ = m λ

In the case of destructive interference we must add half wavelength (λ/2)

   d siyn θ = (m + ½) λ

Let's clear

    d / λ = (m + ½) / sin θ

Let's calculate

   d / λ = (2+ ½) / sin 5.4

   d / λ = 5 / (2 sin 5.4)

   d / λ = 26.7

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In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
Dmitrij [34]

Answer:

\lambda_2 = 573.3 nm

Explanation:

As we know that the position of maximum intensity on the screen is given as

y = \frac{N\lambda L}{d}

here we know that

\lambda = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}

so here we have

8 (645 nm) = 9 \lambda_2

\lambda_2 = 573.3 nm

4 0
3 years ago
If you can’t open the lid of a jelly jar what should you do?
9966 [12]

If I can't open the lid of a jelly jar, I'd keep trying and if I can't open the lid of a jelly jar after the MANY tries I took, I'd ask for help.

6 0
3 years ago
Read 2 more answers
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai
Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

Pb' - Pat = d \times g \times  (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000

Pb' = 122 KPa

3 0
3 years ago
20. Using the picture, how many neutrons does lithium have?
Igoryamba

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

<u>No. of Neutrons = 3 </u>

8 0
3 years ago
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