Question

plz solve this question​

Answer 1

Answer:

Given λ=10pm=100×10−12 m

D= 40 cm = 40 ×10−2 m

β=0.1mm=0.1×10−3 m

β=λDd

d =λDβ

= 100×10−12×40×10−210−3×0.1

= 4 ×10−7 m.

Thank you

Answer 2

Answer:

In the given Question

Wavelength of the monochromatic beam

$\rm \lambda = 100 \: pm$

$\rm \ \: \: \: \: = 100 \times {10}^{ - 12} \:m$

Distance of the screen from the slit is given by

$\rm \: D = 40 \: cm$

$\rm \: \: \: \: \: \: \: = 40 \times {10}^{ - 2} \:m$

Distance between the successive Maxima

$\rm \: \beta = 0.1 \: mm$

$\rm\: \: \: \: \: \: \: = 0.1 \times {10}^{ - 3} m$

We have to find the Distance separation between 2 slits

d = ?

we know,

$\rm \beta = \dfrac{\lambda D}{d}$

$\rm \: d = \dfrac{\lambda D}{ \beta }$

Now putting the values

$\rm \: d = \dfrac{100 \times {10}^{ - 12} \times 40 \times {10}^{ - 2} }{0.1 \times {10}^{ - 3} }$

$\rm \: d = \dfrac{ {10}^{2} \times {10}^{ - 12} \times 4 \times {10}^{1} \times {10}^{ - 2} }{ {10}^{ - 1} \times {10}^{ - 3} }$

$\rm \: d = \dfrac{4 \times {10}^{3} \times {10}^{ - 14} }{ {}^{ } {10}^{ - 4} }$

$\rm \: d = {4 \times {10}^{3} \times {10}^{ - 14} }{ {}^{ } \times {10}^{ 4} }$

$\rm \: d = {4 \times {10}^{7} \times {10}^{ - 14} }{ {}^{ } }$

$\rm \: d = {4 \times {10}^{ - 7} }{ {}^{ } } m$

Hense, The distance between 2 slit is 4× 10^-7 m