One case we can take advantage of in fields such as fiber optics is total internal refraction. Here the light never travels passed the boundary, which means when this first occurs the angle of refraction is 90°. If going from glass to air, at what incident angle would we first see total internal refraction? nε sin е=nдsin 90° A Calculate the angle and compare it to the angle measured using the simulation. Show work here.

consider the following given. A red laser travelled from air (with an index of refraction of nair = 1.00) to medium A then to medium B as shown below.1. Considering that medium B has an index of refraction of nB = 2.42, what is the angle of refraction ϴ3 (when the laser emerged into medium B) if the angle of incidence is ϴ1 = 60o ?a. 0.35ob. 2.79oc. 21.0od. Not enough information2. From the figure, which of the following statements is true across the three media?a. light travels the slowest in airb. light travels faster in medium A than in medium Bc. light travels the fastest in medium Bd. light travels faster in medium A than in air.

What are some differences between brightfield and phase contrast microscopy? Discuss the pros and cons of using each.

Diamond is known for being a particularly brilliant stone. The reason behind this is its peculiarly high index of refraction, n = 2.4. A ray of light within a diamond hits the diamond-air (n = 1) interface. Calculate the critical angle for diamond such that the ray be totally internally reflected. Critical angle for diamond-air interface = 0 Once you have determined this, compare the critical angle to the one in the figure below, which is for a glass-air interface? Knowing that glass has a smaller refractive index than diamond, which material will have a larger critical angle. n2=1 n1 =1.5 39° 90° 41.8° 45° 45° Difference between the cirtical angles between the diamond-air and glass-air interface is (report the absolute value of the difference) 0 What is the significance of this comparison with regards to the brilliance of diamond?

Lloyds mirror experiment was conducted by an incident light of 500nm. The mirror is placed 1cm below the source and the screen is placed at 100m from the source perpendicular to the source. Find the distance of the first dark band above the mirror