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Chapter 35: Q77P (page 1077)

A Newton’s rings apparatus is to be used to determine the radius of curvature of a lens . The radii of the nth and (n+20th)bright rings are found to be 0.162cm and 0.368cm, respectively, in light of wavelength 546nm. Calculate the radius of curvature of the lower surface of the lens.

Short Answer

Expert verified

The radius of the curvature of the lower surface of the lens is 1.0m.

Step by step solution

01

Given data

The radii ofnth=0.162cm

The radii of n+20=0.368cm

Wavelength λ=546nm

02

Definition of newton’s ring

Newton's ring is a phenomenon in which an interference pattern is created by the reflection of light between two surfaces; a spherical surface and an adjacent touching flat surface.

03

Determine the radius of the curvature 

The radius of curvature of the xth ring

Rx=0.162cm

For n+20thRn+20=0.368cm

Let the radius of the curvature of the lower surface of the lens be R.

The value of R is given by as

R=Rn+202-Rx220×λR=0.368×10-2 m2-0.162×10-2 m220×546×10-9 mR=0.9998mR=1.0m

Hence, the radius of the curvature of the lower surface of the lens is 1.0m.

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Most popular questions from this chapter

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.

A camera lens with index of refraction greater than 1.30 is coated with a thin transparent film of index of refraction 1.25 to eliminate by interference the reflection of light at wavelength λ that is incident perpendicularly on the lens. What multiple of λgives the minimum film thickness needed?

Transmission through thin layers. In Fig. 35-43, 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.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of and interfere,r3and r4here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers 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.

In Fig. 35-37, two isotropic point sources S1 and S2 emit identical light waves in phase at wavelengthλ. The sources lie at separation on an x axis, and a light detector is moved in a circle of large radius around the midpoint between them. It detects 30points of zero intensity, including two on the xaxis, one of them to the left of the sources and the other to the right of the sources. What is the value of dλ?

Two rectangular glass plates (n=1.60) are in contact along one edge (fig-35-45) and are separated along the opposite edge . Light with a wavelength of 600 nm is incident perpendicularly onto the top plate. The air between the plates acts as a thin film. Nine dark fringes and eight bright fringes are observed from above the top plate. If the distance between the two plates along the separated edges is increased by 600 nm, how many dark fringes will there then be across the top plate.
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