Question

Describe recombination of white light.​

Answer 1

Explanation:

Solution:

\displaystyle \sin A = 1 - \sin^2 AsinA=1−sin

2

A

\displaystyle \sin A = \cos^2 AsinA=cos

2

A

\displaystyle \sin^2 A = \cos^4 Asin

2

A=cos

4

A

\displaystyle 1 - \cos ^2 A = \cos^4 A1−cos

2

A=cos

4

A

\displaystyle 1 = \cos^4 A + \cos^2 A1=cos

4

A+cos

2

A

\displaystyle 1^3 = (\cos^4 A + \cos^2 A)^31

3

=(cos

4

A+cos

2

A)

3

by formula \displaystyle (a+b)^3(a+b)

3

\displaystyle 1 = \cos^{12} A + 3 \cos^{10} A +3 \cos^8 A + \cos^6 A1=cos

12

A+3cos

10

A+3cos

8

A+cos

6

A

Transverse 1 onto the other side we get the condition as given as in the question. Comparing the variables we get \displaystyle a = 1a=1, \displaystyle b = 3b=3, \displaystyle c = 3c=3, \displaystyle d = 1d=1 hence the value of \displaystyle b +\frac{c}{a}+b = \frac{3 + 3}{1 + 1} = \frac{6}{2} = 3b+

a

c

+b=

1+1

3+3

=

2

6

=3

Answer 2

Answer:

$\huge\fbox\red{A}\fbox\pink{n}\fbox\purple{S}\fbox\green{w}\fbox\blue{E}\fbox\orange{r}$

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Recombination Of White Light-

The recombination of white light is defined as white light being dispersed in to seven different colours and gained back with its originality. ... Hence the white light is regained back. This process is called as recombination form of white light.

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Explanation:

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