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Aleksandr [31]
3 years ago
11

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

een the first and second minima in the diffraction pattern is 4.75 mm. What is the width (in mm) of the slit?
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

0.2 mm

Explanation:

As we know that

Y = \frac{m\lambda * D}{d}

where

m represents  the order of minimum

y represents the  distance on the screen of the minimum from central axis

λ is the wavelength  of the light

D is the distance between  screen-to-slit

d represents the width of the slit

For first minima

y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}

For second minima

y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}

Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm

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A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field
stepladder [879]

Answer:

\vec{E} =  1.2\times 10^5(-i)

Explanation:

Given that

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F = 0

0=q.\vec{E}+ q.(\vec{v}\times \vec{B})

0=\vec{E}+ (\vec{v}\times \vec{B})

0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)

\vec{E} =- 2\times 10^5(-j) \times 0.6(-k)

\vec{E} =-1.2\times 10^5(i)

\vec{E} = 1.2\times 10^5(-i)

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