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jek_recluse [69]
3 years ago
13

Light from a helium-neon laser (λ = 633 nm) illuminates two slits spaced 0.50 mm apart. A viewing screen is 2.5 m behind the sli

ts. What is the spacing between two adjacent bright fringes?
Engineering
1 answer:
serg [7]3 years ago
7 0

Answer:

Explanation:

Given

wavelength \lambda =633\ nm

distance between two slits d=0.5\ mm

Screen is placed at a distance L=2.5\ m

Location of a (n+1)th bright fringe is given by

x_{n+1}=\frac{L}{2d}(n+1)\lambda

for nth bright fringe

x_n=\frac{L}{2d}(n)\lambda

Distance between two bright fringes

x_{n+1}-x_n=\frac{L}{2d}\cdot \lambda

x_{n+1}-x_n=\frac{2.5}{2\times 0.5\times 10^{-3}}\cdot 633\times 10^{-3}

x_{n+1}-x_n=1.582\ mm

You might be interested in
Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output tho
Likurg_2 [28]

Answer:

See explanation

Explanation:

//Include the

//required header files.

#include <stdio.h>

//Define the

//main() function.

int main(void) {

//Declare the

//required variables.

char input_char;

int input_int;

double input_double;

char input_string[100];

//Prompt the user

//to enter an integer.

printf("Enter integer: ");

//Read and store

//the integer.

scanf("%d", &input_int);

//Prompt the user

//to enter a double value.

printf("Enter double: ");

//Read and store

//the double value.

scanf("%lf", &input_double);

//Prompt the user

//to enter a character.

printf("Enter character: ");

//Read and store

//the character.

scanf(" %c", &input_char);

//Prompt user to

//enter the string

printf("Enter string: ");

//Read and

//store the string.

scanf("%s", input_string);

//(1)

//Display the values.

printf("%d %lf %c %s\n",

input_int, input_double,

input_char, input_string);

//(2)

//Display the values

//in reverse order.

printf("%s %c %lf %d\n",

input_string, input_char,

input_double, input_int);

//(3)

//Cast the double to

//an integer and display it.

printf("%lf cast to an integer is %d",

input_double, (int)(input_double));

//Return from the

//main() function.

return 0;

}

4 0
3 years ago
How might constraints and trade-offs affect the final design of a bridge?
Rainbow [258]

Answer:

It will limit the materials for your bridge which could cause damage to your bridge

Explanation:

4 0
3 years ago
Read 2 more answers
When you see a street with white markings only, what kind of street is it?
Georgia [21]

Answer:

it's a one way street

3 0
4 years ago
1. The sine rule is used when we are given either a) two angles and one side, or b) two sides and a non-included angle.
sammy [17]

Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

Or

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

3 0
3 years ago
a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

5 0
4 years ago
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