What is Brewster’s angle for an air-glass (n = 1.56) surface? Specify two answers.

Light of wavelength 5.0 x 10⁻⁷ passes through two parallel slits and falls on a screen 5.0 m away. Adjacent bright bands of the interference pattern are 2.0 cm apart.

(a) Find the distance between the slits.

(b) The same two slits are next illuminated by light of a different wavelength, and the fifth minimum for this light occurs at the same point on the screen as the fourth minimum for the previous light. What is the wavelength of the second source of light?

A single optical coating reduces reflection to zero for λ = 550 nm. By what factor is the intensity reduced by the coating for λ = 430 nm and λ = 670 nm as compared to no coating? Assume normal incidence.

Show that the radius r of the mᵗʰ dark Newton’s ring, as viewed from directly above (Fig. 34–18), is given by r = √mλR where R is the radius of curvature of the curved glass surface and λ is the wavelength of light used. Assume that the thickness of the air gap is much less than R at all points and that r ≪ R . [Hint: Use the binomial expansion.]

What would Brewster’s angle be for reflections off the surface of water for light coming from beneath the surface? Compare to the angle for total internal reflection, and to Brewster’s angle from above the surface.

Light of wavelength 690 nm passes through two narrow slits 0.66 mm apart. The screen is 1.75 m away. A second source of unknown wavelength produces its second-order fringe 1.23 mm closer to the central maximum than the 690-nm light. What is the wavelength of the unknown light?

(II) When a Newton’s ring apparatus (Fig. 34–18) is immersed in a liquid, the diameter of the tenth dark ring decreases from 2.92 cm to 2.54 cm. What is the refractive index of the liquid? [Hint: See Problem 37.]