Minimum Deviation of a Prism

Introduction to minimum deviation of a prism:
A prism is a transparent medium bounded by two plane surfaces inclined to each other at a suitable angle. The angle between two surfaces is known as refracting angle or angle of prism. In a prism, a ray of light suffers two refraction and the result is deviation. In other words, we say that after passing through prism the ray of light deviates through a certain angle from its original path.

Minimum Deviation of a Prism:

Look at the given diagram, here ABC is the principal section of the prism and the angle of prism is A.

A ray of light KL is incident on face AB of the prism at Ði1. It bends towards the normal NO and refracted along LM at Ðr1. The refracted ray LM is incident at Ðr2 on face AC of the prism. It bends away from the normal PO and emerges MS at angle Ði2. In passing through the prism, the ray KL suffers two refraction and finally turned through an ÐQTM = d, this is the angle of deviation.


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Calculation for the Minimum Deviation of a Prism:
In  TLM, `delta` = TLM + ÐTML

`delta` = (i1 – r1) + (i2 – r2)

`delta` = (i1 + i2) - (r1 + r2)                       ….(1)

In D OLM, ÐO + r1 + r2 = 180°           ….(2)

In quadrilateral ALOM,

ÐL + ÐM = 180°

ÐA + ÐO = 180°

Put this value in equation (2), we get

ÐO + r1 + r2 = ÐA + ÐO

r1 + r2 = ÐA                                                     ….(3)

Put this value in equation (1)

`delta` = (i1 + i2) – A                                                ….(4)

Let n be the refractive index of the medium of prism with respect to air

n = Sin i1 / Sin r1 = i1 /r1                        (when angles are small)

i1 = n r1 , Similarly i2 = n r2

Put this value in equation (4), we get

d = n(r1 + r2) – A

d = n A – A

`delta` = (n – 1)A

In case of minimum deviation of a prism, i1 = i2 = i so that r1 = r2 = r, so put these values in equation (3) and in equation (4), we get

r + r = A

r = A/2

`delta`m = (i + i) – A = 2i – A   where dm is the minimum deviation angle

i = ( `delta`m + A)/2

n =  

Plotting Ordered Pairs

Introduction to plotting ordered pairs:

Plotting ordered pairs mean we have the x and y value in a graph.  It is denoted by the open and close parenthesis. Let us take an example for plotting the ordered pairs (3, 4). In this 3 is the x axis value and 4 is the y axis value. X axis value it represent the x component length and the y value mean it represent the y component length. If we placed an ordered pair in a grid it is called coordinates.

Method to Plotting Ordered Pairs:

We will plot the ordered pairs on the graph based on the origin. In a graph the origin is also a ordered pair which is (0, 0). Here the first 0 indicates the x coordinate and the second 0 indicates the y coordinate. Always the x coordinate is indicated first. So if we want to place any coordinates we have to horizontally and vertically from the origin. If we want to locate the positive x value we have to place it left to the origin. If we want to place the negative x value we have to move right side of the origin. Likewise if we want to place the positive y value we have to move upward and for negative y value we have to move downwards of the origin. Let us see some examples for plotting ordered pairs on a graph.

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Examples for Plotting Ordered Pairs:

Ex:1 Plot the given ordered pairs (2, 2), (2, 3), (5, 6)

Sol:

Let us take the first ordered pair. To plot the ordered pairs we have to move 2 units horizontally and 2 units vertically.

Here the x component value is 2 and the y component value is 2.

Likewise we have to plot the second and the third ordered pairs. For the second ordered pair we have to move 2 units horizontally and 3 units vertically. For the third ordered pair we have to move 5 units horizontally and 6 units vertically.

Solving discrete variables

Introduction:

      In our article let us discuss the solving discrete variables. Solving discrete variables is the important one in math and it explains how to solve discrete variable. Discrete variables are mainly  based on two types of variables and we will study about solving discrete variables with its example.

Example for discrete variable: In an event of the nearest millisecond to the time.

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Solving Discrete Variable:

 Description about solving discrete variable:

Solving discrete variable:

    Solving discrete variable is the important one in mathematics. The discrete variable only takes the values under in its boundary itself. The limits of discrete value are completed. The best example for discrete variable is 'Real world quantities approximation'.

Example for solving discrete variable:

Solution:

            If the student flips the coin, he counts how much possible number of heads as outcomes in an  event. The possible number of head values is present between 0 and other values such as x=1, 2, 3 like that.

             The above examples are used to solve the discrete variable with its discrete probability function.

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Solving Discrete Variable Probability Distribution:-

DISCRETE PROBABILITY DISTRIBUTION:-

Definition:

If the variable is said to be discrete variable then its probability distribution of the variable is known as discrete probability distribution.

The following example clearly explains the concept of discretr random variable with its probability function.

For example:

solution:

If a coin flips in twice, we get  4 possible values of outcomes such as HH, HT, TH, and TT.

The random variable X represents number of heads in an event.



                                  No. of heads                                     Probability

                                        0                                                        `1/4`

                                        1                                                        `1/2`

                                        2                                                        `1/2`

Another example:

An event of flipping a coin three times then find the probability of possible number of tails as outcomes and x=1,2,3 ?

soluton:

If you flip the coin 3 times, we shall get possible values of 8 outcomes such as HHH, HHT, HTH, HTT, THH, THT, TTH and TTT

Here the possible number of values for the above event is showed below.

                                  Number of tails                                  Probability

                                             0                                              `1/8`

                                             1                                               `7/8`

                                             2                                               `5/8`

                                             3                                              `1/8`

so the above examples clears the concept of solving discrete variable.

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Introduction to adding and subtracting of matrices

Introduction to adding and subtracting of matrices:
In this article let us learn what is a matrix and how to add and subtract matrices.

What is a matrix:

Consider a set of entries or elements. When they are arranged in rows and columns enclosing them with a square bracket [ ] or parenthesis ( ) is called as matrix.

We can denote matrix using capital letters like A, B, C...

The set of entries or elements in the matrix could be real or complex numbers, algebraic expressions.

In matrix, Counting the horizontal segment from top to bottom is termed as number of rows. 

                Counting the vertical segment from left to right is termed as number of columns.

Example:

R1,R2,R3 represent rows

C1,C2,C3 represent columns.

Rules to be Followed while Adding and Subtracting Matrices:

 When the matrices have same number of rows and columns, then we can add them.
       For example:

        Any two matrices of 5 rows and 3 columns could be added.

        When a  matrix with 2 rows and 3 columns and a matrix with 3 rows and 4 columns could not be added.

 Add the corresponding rows and columns of the given two matrix and the resulatat matrix is the added matrix.
      Example:

      We could add or subtract the element in the third row and second column of the first matrix  with the element in the third row and second column of the second matrix and not with any other entries. that is , corresponding entries are added or subtracted. For example  the element in the first row first column of the first matrix can be  added to the element in the first row first column in the second matrix.

Subtraction is nothing but  negative addition.
Additive inverse of matrix A is - A
             A + ( -A ) = ( -A ) + A =  zero matrix

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Problems on Adding and Subtracting Matrices:

Let us  add and subtract the given matrices

A = `[[2,5,7],[0,8,1]]` : B = `[[6,3,6],[0,3,5]]`

A+B = `[[2+6,5+3,7+6],[0+0,8+3,1+5]]`

         = `[[8,8,13],[0,11,6]]`

A - B = `[[2-6,5-3,7-6],[0-0,8-3,1-5]]`

         = `[[-4,2,1],[0,5,-4]]`

Introduction to solving discrete variables

Introduction to solving discrete variables:

      In our article let us discuss the solving discrete variables. Solving discrete variables is the important one in math and it expailns how to solve discrete variable. Discrete variables are mainly  based on two types of variables and we will study about solving discrete variables with its example.
Example for discrete variable: In an event of the nearest millisecond to the time.

Solving Discrete Variable:

Description about solving discrete variable:

Solving discrete variable:

    Solving discrete variable is the important one in mathematics. The discrete variable only takes the values under in its boundary itself. The limits of discrete value are completed. The best example for discrete variable is 'Real world quantities approximation'.

Example for solving discrete variable:

Solution:

            If the student flips the coin, he counts how much possible number of heads as outcomes in an  event. The possible number of head values is present between 0 and other values such as x=1, 2, 3 like that.

             The above examples are used to solve the discrete variable with its discrete probability function.

Solving Discrete Variable Probability Distribution:-

DISCRETE PROBABILITY DISTRIBUTION:-

Definition:

If the variable is said to be discrete variable then its probability distribution of the variable is known as discrete probability distribution.

The following example clearly explains the concept of discretr random variable with its probability function.

For example:

solution:

If a coin flips in twice, we get  4 possible values of outcomes such as HH, HT, TH, and TT.

The random variable X represents number of heads in an event.

                                  No. of heads                                     Probability

                                        0                                                        `1/4`

                                        1                                                        `1/2`

                                        2                                                        `1/2`

Another example:

An event of flipping a coin three times then find the probability of possible number of tails as outcomes and x=1,2,3 ?

soluton:

If you flip the coin 3 times, we shall get possible values of 8 outcomes such as HHH, HHT, HTH, HTT, THH, THT, TTH and TTT

Here the possible number of values for the above event is showed below.

                                  Number of tails                                  Probability

                                             0                                              `1/8`

                                             1                                               `7/8`

                                             2                                               `5/8`

                                             3                                              `1/8`

so the above examples clears the concept of solving discrete variable.

Definition of intersecting lines

Definition of intersecting lines:

Consider the equations

a₁x+b₁y+c₁ = 0

a₂x+b₂y+c₂ =0

These equations represent two straight lines. If a point P lies either in  any equation then P satisfies the equation

(a₁x+b₁y+c₁ = 0) (a₂x+b₂y+c₂) =0→ (3)

Conversely if P is a point satisfying the equation (3), then P satisfies either (1) or (2) and hence p lies either (1) or (2). Thus equation (3) represents both the lines (1) and (2).By expanding the equation (3) it can be reduced to the form ax²+2hxy+by²+2gx+2fy+c=0 a second degree linear equation in x and y.

            Consequently the combined equation of the two lines a a₁x+b₁y=0, a₂x+b₂y=0 which passes through the origin is (a₁x+b₁y)(a₂x+b₂y) =0 and it is of the form ax²+2hxy+by² = 0, a second degree homogeneous equation in x and y.

1)If h²≥ ab, then ax²+2hxy+by² = 0 represents a pair of straight lines passing through the origin.

Note:- If h²< ab then ax²+2hxy+by² = 0 represents two imaginary lines having real  point of intersection, the origin.

Note 2:- If h²= ab then ax²+2hxy+by² = 0 represents coincident lines.

Note 3:- h²> ab then ax²+2hxy+by² = 0 represents two real and different lines.

Intersecting lines:- If the two pair of lines are intersects at a point is called the pair of intersecting lines.

Angle between a pair of lines:- If ө is an angle between the lines represented by ax²+2hxy+by² = 0 then cos ө = (a+b)/√((a-b)²+4h²

1)If ө be the acute angle between the lines represented by ax²+2hxy+by² = 0, then cos ө =( mod of a+b)/ √((a-b)²+4h²

2) If ө be the acute angle between the lines represented by ax²+2hxy+by² = 0, then Tan ө = 2√(h²-ab)

3) If ө=0, then the two lines will be coincident and tanө = 0 which implies h²-ab =0  which implies h²=ab

3)If ө= ∏ /2, then the two lines will be perpendicular and cos ө = 0 which implies a+b =0 which implies co efficient of x²+coefficient of y²

More on Definition of Intersecting Lines 

Angle bisectors:- The equations of bisectors of angle between the lines a₁x+b₁y+c₁ = 0, a₂x+b₂y+c₂ =0 are

(a₁x+b₁y+c₁)/ √(a₁²+b₁²) (+ or -) (a₂x+b₂y+c₂ )/ √(a₂²+b₂²) =0 If c₁c₂>0 then the line bisection the angle containing the origin between the lines is

(a₁x+b₁y+c₁)/ √(a₁²+b₁²)  - (a₂x+b₂y+c₂ )/ √(a₂²+b₂²) =0 and the line bisecting  the other angle is

(a₁x+b₁y+c₁)/ √(a₁²+b₁²) + (a₂x+b₂y+c₂ )/ √(a₂²+b₂²) =0

Pair of angle bisectors:- The equation to the pair of bisectors between the pair of  lines ax²+2hxy+by² = 0 is h(x²-y²) = (a-b)xy

DEFINITION:-A pair of lines L₁L₂ = 0 is said to be equally inclined to a line L =0 if the lines L₁ =0, and L₂=0 subtend the same angle with the line L =0

1)The equation to the pair of lines passing through the origin and forming an equilateral triangle with the line ax+by+c=0 is (ax+by)²- 3(bx-ay)²=0

Solved Problems on Definition of Intersecting Lines

Q 1: Find the equations of bisectors of the angle between the lines x+y-2 =0 and x-7y+5 =0 and distinguish them.

Sol :- Equations of the bisectors of the angle between the given lines are   (x+y-2) / √(1+1) (+ or -) (x-7y+5)/ √(1+49) =0

          (x+y-2) / √2 (+ or -) (x-7y+5)/ 5√2 = 0

        5(x+y-2)(+ or -) 1(x-7y+5) = 0

5x+5y-10+x-7y+5 = 0 or 5x+5y-10-x+7y-5 = 0

6x-2y-5 =0 or 4x+12y-15 = 0

If ө is the acute angle between x+y-2 =0, 6x-2y-5 =0 then

            Cos ө = mod of (6-2)/( √(1+1) √(36+4) = 4/4√5 = 1/√5<1/√2

                                                                                         = cos (∏ /4)

                                                                                          = ө >∏ /4

Therefore 6x-2y-5 =0 is the obtuse angle bisector and hence 4x+12y-15 =0 is the acute angle bisector.

Q 2:  Find the equation to the pair of lines passing through the origin and forming an equilateral triangle with the line 2x+3y+5 =0 Also find the area.

Sol :- The equation to the required parallel lines is

       (2x+3y)²-3(3x-2y)² =0

   (4x²+9y²+12xy)-3(9x²+4y²-12xy) = 0

   -23x²+48xy-3y² = 0

   23x²-48xy+3y² = 0

Therefore the length of altitude p = 5/ √(4+9) = 5/√(13)

Area of the triangle = P²/√3 = 25/13√3 sq. unit

Math Problems Help

Introduction to solving factoring calculator:

In this article we are discussing about solving factoring calculator. The decomposition of the elements is called as factors of that element. We can get the elements by multiplying the some of its factors. For example the number 12 can be factored into 3 x 4 and x² - 16 can be factored into (x-4) (x+4).The factors can be positive elements. The devices that use to get an output by giving proper input are known as calculators. By using the solving factoring calculator we can solve for factors of the number.

Solving Factoring Calculator - Example Problems:
Example 1:
Factor the given expression, 2x² + 4x
Solution:
= 2x² + 4x
Taking the common terms outside,
= 2x(x + 2)
The answer is 2x (x+2).
Example 2:
Factor the given expression, 5x² + 155x = 0
Solution:
5x² + 155x = 0
Taking the common terms outside,
5x(x + 31) = 0
5x = 0
Divide 5 on both sides,

x = 0
x + 31 = 0
Subtract 31 on both sides,
x + 31 – 31 = 0 – 31
x = -31
The answer is 0,-31.