A parallel monochromatic light (=600nm) falls normally on a circular aperture of diameter 0.12¢m and is viewed from the opposite side along a line through the center of the hole and normal to its plane. Calculate the three larger distances from the hole at which the intensity is zero.
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- Problem 1: You want to test a special watertight monochromatic light source in an un- derwater environment. If the light source shines its beam of electromagnetic radiation into your eyes when you and the light source are above the water, with both your eyes and the light source immersed in air, the beam appears to your eyes to have a deep violet color. The index of refraction of water is about 1.33 What color, if any, would your eyes perceive the beam to have when you and the light source are submerged in water – and why? (A) No color. The beam would become invisible under water because its wavelength in water is shorter than in air, corresponding to a wavelength of ultra-violet light. The beam's wavelength in the water determines its visibility to the eye under water, and ultra-violet light is not visible to the human eye. (B) Same color as in air, above the water, i.e., deep violet. Under water, the beam would have the same wavelength as in air. The beam's wavelength in the water…Plz do fast asapWhich of the following statements are (or could be) true? Choose all that apply. O A glass at a temperature of 43 K will emit EM waves. O The index of refraction of a newly discovered transparent material is 1.6. In a vacuum, radio waves move the same speed as microwaves. O Total internal reflection will occur if the critical angle is greater than the incident angle. A light ray was reflected. The incident angle was 44° and the reflected angle was 60°. O If do = 0, then d; = f.
- The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just‑resolvable points of light for a 6.25 mm diameter pupil, assuming the average wavelength of 539 nm? angle between two points of light: ° Take the result to be the practical limit for the eye. What is the greatest possible distance a car can be from a person if he or she can resolve its two headlights, given they are 1.20 m apart? greatest distance at which headlights can be distinguished: m What is the distance between two just‑resolvable points held at an arm's length (0.750 m) from a person's eye? distance between two points 0.750 m from a person's eye:White light is sent through an interface of a 100% (w/v) glycerol solution (n = 1.474) and a 20% (w/v) sucrose solution (n2 = 1.364) at an angle of Oj. Incident ray Reflocted ray If O, = 33', determine the angle of O2 in degrees. Refructive index = , O2 Refractive index – n, If O, = 0, determine the angle of 02 in degrees. Refractod rayCalculate the angle of refraction at the air/core interface, r . critical angle, c and incident angle at the core/cladding interface, i . Will this light ray propagate down the fiber? You have the following data: nair = 1, ncore = 1.46, ncladding =1.43, incident =12o Answers: r = 8.2o , c = 78.4o , i = 81.8o light will propagate Formula summery Index of Refraction n= n sin 01= n, sin 0, o = sin1(2) Snell's Law Critical Angle Acceptance angle a= sin1 Numerical Aperture NA = sin a = ni- nị