If the distance between the first and tenth minima of a double-slit pattern is 18.0 mm and the slits are separated by 0.150 mm with the screen 50.0 cm from the slits, what is the wavelength of the light used?

Short Answer

Expert verified

Thus, the wavelength of light is 600nm.

Step by step solution

01

The separation of slits.

The condition for a minimum in the two-slit interference pattern isdsinθ=(m+12)λ,where dis the slit separation, λis the wavelength,mis an integer, and θis the angle made by the interfering rays with the forward direction. If θis small, sinθmay be approximately byθin radians.

Then, , θ=m+12λdand the distance from the minimum to the central fringe is:

y=DtanθDsinθDθ=m+12Dλd

Where Dis the distance from the slits to the screen. For the first minimumm=0 for the tenth onem=9 . The separation is:

Δy=9+12Dλd-12Dλd=9Dλd

02

Calculate the wavelength of light. 

Solve for the wavelength as follows:

λ=dΔy9D=0.15×10-1m18×10-1m950×10-2m=6.0×10-7m=600nm

Hence, the wavelength of light is 600nm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In two experiments, light is to be sent along the two paths shown in Fig. 35-35 by reflecting it from the various flat surfaces shown. In the first experiment, rays 1 and2 are initially in phase and have a wavelength of 620.0nm. In the second experiment, rays 1 and2 are initially in phase and have a wavelength of 496.0nm . What least value of distance L is required such that the 620.0nmwaves emerge from the region exactly in phase but the 496.0nmwaves emerge exactly out of phase?

Figure 35-25 shows two sources s1 and s2 that emit radio waves of wavelengthλin all directions. The sources are exactly in phase and are separated by a distance equal to 1.5λ . The vertical broken line is the perpendicular bisector of the distance between the sources.

(a) If we start at the indicated start point and travel along path 1, does the interference produce a maximum all along the path, a minimum all along the path, or alternating maxima and minima? Repeat for

(b) path 2 (along an axis through the sources) and

(c) path 3 (along a perpendicular to that axis).

In Fig. 35-4, assume that the two light waves, of wavelength 620nm in air, are initially out of phase by π rad. The indexes of refraction of the media are n1=1.45 andn2=1.65 . What are the (a) smallest and (b) second smallest value of Lthat will put the waves exactly in phase once they pass through the two media?

A thin film of liquid is held in a horizontal circular ring, with air on both sides of the film. A beam of light at wavelength 550 nm is directed perpendicularly onto the film, and the intensity I of its reflection is monitored. Figure 35-47 gives intensity I as a function of time the horizontal scale is set by ts=20.0s. The intensity changes because of evaporation from the two sides of the film. Assume that the film is flat and has parallel sides, a radius of 1.80cm, and an index of refraction of 1.40. Also assume that the film’s volume decreases at a constant rate. Find that rate.

Reflection by thin layers. In Fig. 35-42, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1, n2, and n3, the type of interference, the thin-layer thickness in nanometres, and the wavelength λ in nanometres of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free