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Chapter 35: Q77P (page 1077)

A Newton鈥檚 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 位=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脳位R=0.368脳10-2鈥�m2-0.162脳10-2鈥�m220脳546脳10-9鈥�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 with index of refraction n=1.40 is placed in one arm of a Michelson interferometer, perpendicular to the optical path. If this causes a shift of 7.0 bright fringes of the pattern produced by light of wavelength 589nm, what is the film thickness?

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, n2and n3, the type of interference, the thin-layer thickness Lin nanometres, and the wavelength 位in nanometres of the light as measured in air. Where 位is missing, give the wavelength that is in the visible range. Where localid="1663142040666" Lis missing, give the second least thickness or the third least thickness as indicated


Does the spacing between fringes in a two-slit interference pattern increase, decrease, or stay the same if

(a) the slit separation is increased,

(b) the color of the light is switched from red to blue, and

(c) the whole apparatus is submerged in cooking sherry?

(d) If the slits are illuminated with white light, then at any side maximum, does the blue component or the red component peak closer to the central maximum?

Two parallel slits are illuminated with monochromatic light of wavelength 500 nm. An interference pattern is formed on a screen some distance from the slits, and the fourth dark band is located 1.68 cm from the central bright band on the screen. (a) What is the path length difference corresponding to the fourth dark band? (b) What is the distance on the screen between the central bright band and the first bright band on either side of the central band? (hint: The angle to the fourth dark band and the angle to the first band are small enough that 迟补苍胃鈮�蝉颈苍胃)

In Fig. 35-45, a broad beam of light of wavelength 620 nm is sent directly downward through the top plate of a pair of glass plates touching at the left end. The air between the plates acts as a thin film, and an interference pattern can be seen from above the plates. Initially, a dark fringe lies at the left end, a bright fringe lies at the right end, and nine dark fringes lie between those two end fringes. The plates are then very gradually squeezed together at a constant rate to decrease the angle between them. As a result, the fringe at the right side changes between being bright to being dark every 15.0 s.

(a) At what rate is the spacing between the plates at the right end being changed?

(b) By how much has the spacing there changed when both left and right ends have a dark fringe and there are five dark fringes between them?
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