Answer

Verified

417.9k+ views

**Hint:**First consider that ${{n}^{th}}$ minima of first wavelength coincides with ${{m}^{th}}$ minima of second wavelength. Then equate both the equations. Then find the value of m and n in two cases considering two ratios. Then the location of minima is calculated using both the values of m and n. Thus the minimum distance between two successive regions is the difference of those two distances.

**Formula used:**

$\dfrac{(2n-1){{\lambda }_{1}}D}{2d}=\dfrac{(2m-1){{\lambda }_{2}}D}{2d}$

where, d is the distance between the slits.

D , the distance between the slit and screen.

${{\lambda }_{1}}\And {{\lambda }_{2}}$ are the wavelengths of light used.

m and n are the order of interference.

**Complete answer:**

Let ${{n}^{th}}$ minima of 400nm coincides with ${{m}^{th}}$ minima then,

$\dfrac{(2n-1){{\lambda }_{1}}D}{2d}=\dfrac{(2m-1){{\lambda }_{2}}D}{2d}$

$\Rightarrow $ (2n-1)$\times 400=(2m-1)\times 560$

$\dfrac{2n-1}{2m-1}=\dfrac{560}{400}=\dfrac{28}{20}=\dfrac{14}{10}=\dfrac{7}{5}$ =$\dfrac{21}{15}$

If we the ratio $\dfrac{7}{5}$, n=4 and m=3. Then take the second ratio $\dfrac{14}{10}$ , then n=7.5 and m=5.5. This is not acceptable. So consider the next ratio $\dfrac{21}{15}$ , n=11 and m= 8.

That is ${{4}^{th}}$ minima of 400nm coincides with the ${{3}^{rd}}$ minima of 560nm.Then the location of minima is,

${{y}_{1}}=\dfrac{(2\times 4-1)\times 400\times {{10}^{-6}}\times 1000}{2\times 0.1}=14mm$

Similarly ${{11}^{th}}$ minima of 400nm coincides with ${{8}^{th}}$ minima of 560nm. Then the location of minima is,

${{y}_{2}}=\dfrac{(2\times 8-1)(560\times {{10}^{-6}})\times 1000}{2\times 0.1}=42mm$

Therefore the required distance is,

${{y}_{2}}-{{y}_{_{1}}}=42-14=28mm$

**Then the option(4) is correct.**

**Additional information:**

To obtain constructive interference for a double slit, the path length difference must be an integral multiple of the wavelength.

$d\sin \theta =m\lambda $

For m= 0,1,-1,2,-2,…… (Constructive interference)

$d\sin \theta =\left( m+\dfrac{1}{2} \right)\lambda $

For m=0,1,-1,2,-2,……….(Destructive interference)

Here, $\lambda $ is the wavelength of light, d is the distance between the slits and m is the order of interference.

**Note:**

The value of m and n should not be a decimal or a fractional number. While calculating the value of m and n if we are getting a fractional or decimal number then we should take another set of ratios and we should find the values of m and n. Then by taking two values of m and n we can find the location of minima and the required distance is their difference.

Recently Updated Pages

Which of the following is correct regarding the Indian class 10 social science CBSE

Who was the first sultan of delhi to issue regular class 10 social science CBSE

The Nagarjuna Sagar project was constructed on the class 10 social science CBSE

Which one of the following countries is the largest class 10 social science CBSE

What is Biosphere class 10 social science CBSE

Read the following statement and choose the best possible class 10 social science CBSE

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write a letter to the principal requesting him to grant class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

10 examples of law on inertia in our daily life