Question

Suppose Young's experiment performed with blue-green light of wavelength 500 nm. The
slits are 1.2 mm apart and the viewing screen is 5.4 m from the slits. How far apart are the
bright fringes?​

Answer 1

Given

Suppose that Young's experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits

To Find

How far apart are the bright fringes near the center of the interference pattern

Solution

We know that

$\sf \longrightarrow \Delta y=\dfrac{\lambda D}{d}$

According to the question

We are asked to find "Δy"

Given that

Suppose that Young's experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits

Hence

• λ = 500 nm
• D = 5.4 m
• d = 1.2 mm

Substituting the values

We get

$\sf \longrightarrow \Delta y=\dfrac{500 \times 10^{-9} \times 5.4}{1.2 \times 10^{-3}} \ m$

On further simplification

We get

$\sf \longrightarrow \Delta y=2.25 \times 10^{-3} \ m$

Therefore

$\sf \longrightarrow \Delta y=2.25 \ mm$

Answer 2

$\huge {\blue {\underline {\underline {\mathtt{Question : }}}}}$

Suppose that Young's experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits.

How far apart are the bright fringes near the center of the interference pattern.

$\huge {\red {\underline {\underline {\mathtt{Answer : }}}}}$

We know that ,

$\sf \longrightarrow \Delta y=\dfrac{\lambda D}{d}$

Given :

Suppose that Young's experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart, and the viewing screen is 5.40 m from the slits.

To find :

$\sf value\: of \: \Delta y$

Solution :

• λ = 500 nm
• D = 5.4 m
• d = 1.2 mm

Substituting the values

We get

$\sf : \implies \Delta y=\dfrac{500 \times 10^{-9} \times 5.4}{1.2 \times 10^{-3}}$

$\therefore \sf \Delta y=2.25 \times 10^{-3}$

$\boxed{\sf \Delta y=2.25 \ mm}$