A set of narrow vertical slits is located a distance D from a screen. The slits

Question

A set of narrow vertical slits is located a distance D from a screen. The slits are equally spaced and have the same width. The intensity pattern in the figure is observed when light from a laser passes through the slits, illuminating them uniformly. The screen is perpendicular to the direction of the light. Data: Distance to the screen = 2.89 m, Wavelength of light = 470 nm, Distance between tick marks on the intensity figure = 1.50 cm.

What is the spacing between the slits? I got this as 9.24E-5 m. (true).

Calculate the width of the slits.

If the slits separation is increased by a factor of 4, what would be the distance between the principle peaks on the screen?

A set of narrow vertical slits is located a distance D from a screen. The slits are equally spaced and have the same width. The intensity pattern in the figure is observed when light from a laser passes through the slits, illuminating them uniformly. The screen is perpendicular to the the Intensity flaure 1.50 cm. through her ots, lifurminating thernt uniformniy The screen is perpen dieular to the direct.on of the lig ht Data: Distanve to the s cneensit 2as n Wavrel engur of ight 470 h istatween tick marks on ion of the light. Data: Distance to the screen-2.89 m, wavelength of light = 470 nm, Distance betweentick marks on What is the spacing between the slits? 9.240×10-3 m You are correct. Your receipt no. is 160-1084 Previous Tres Calculate the w61084 Calculate the width of the slits. 1.535-5 m There is a fixed relation between wavelength, slitwidth, and angle between the first minimum due to the slit width and the central maximum. There is a difference between the minima due to the slit to slit Interference and the minima due to the diffraction due to the slit widths. Compare with the fiqure of the prevlous problem. The angle can be detenmined from the distance to the screen and the separation between the first minimum and the central maximum on the screen. SulmeIncorrect. Tries 1/12 Previous Tries f the slits tion is increased by a factor of 4, what would be the distance between the principal peaks on the screen? Submt AneTries 0/12

Explanation / Answer

given
distance from the screen, D = 2.89 m
waveength of light, lambda = 470 nm
First maxima's position, y = 1.5 cm
so for first maxima in multi slit interference, of slit seperation a is
a*sin(theta) = lambda
sin(theta) = y/D
a = 470*10^-9*2.89/1.5*10^-2 = 9.055*10^-5 m [ spacing between the slits]

now, ratio of slit width to slit seperation = d/a = 1 ( from the figure, as number of minimas between 2 secondary maximas iwhen divided by 2 gives this)
so, d = 9.955*10^-5 m

if slit seperastion is increased by factor of 4,
a = 4d
so, the tw0 principal maximas will have one wavelet between principal maximas on the graph

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