Physics, 06.05.2020 20:59 memoryofdale

A single slit forms a diffraction pattern, with the first minimum at an angle of 40 degrees from central maximum. Monochromatic light of 410 nm wavelength is used. The same slit, illuminated by a different monochromatic light source, produces a diffraction pattern with the second minimum at a 60 degree angle from the central maximum. The wavelength of this light, in nm, is closest to:

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Answer from: 12camtheman

276.19nm

Explanation:

To find the other wavelength you use the following condition for the diffraction of both wavelengths:

$m_1\lambda_1=asin\theta_1\\\\m_2\lambda_2=asin\theta_2\\\\$ ( 1 )

λ1=410nm

m=1 for wavelength 1

m=2 for wavelength 2

a: width of the slit

θ1: angle of the first minimum

θ2: angle of the second minimum

you divide both equations and you obtain:

$\frac{m_1\lambda_1}{m_2\lambda_2}=\frac{sin\theta_1}{sin\theta_2}\\\\\lambda_2=\frac{sin\theta_2}{sin\theta_1}\frac{m_1\lambda_1}{m_2}\\\\\lambda_2=\frac{sin60\°}{sin40\°} \frac{(1)(410nm)}{2}=276.19nm$

hence, the wavelength of the second monochromatic wave is 276.19nm

Answer from: amandajbrewerdavis

The wavelength of the second monochromatic light is $\lambda = 277 nm$

Explanation:

From the question we are told that

The angle of the first minimum is $\theta_1 = 40^o$

The wavelength of the first monochromatic light is $\lambda_1 = 410 \ nm$

The angle of the second minima is $\theta_2 = 60^o$

For the first minima the distance of separation of diffraction patterns is mathematically represented as

$a = \frac{\lambda_1 }{sin \theta_1}$

Substituting values

$a = \frac{410 *10^{-9}}{sin (40) }$

a = 638 nm

The distance between two successive diffraction is constant for the same slit

Thus the wavelength of the second light is

$\lambda = \frac{a * sin (60)}{2}$

Substituting value

$\lambda = \frac{638 * sin (60)}{2}$

$\lambda = 277 nm$

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A single slit forms a diffraction pattern, with the first minimum at an angle of 40 degrees from cen...