Electric room heaters use a concave mirror to reflect infrared (IR) radiation from hot coils. Note that IR radiation follows the same law of reflection as visible light. The mirror has a radius of curvature of 40 cm and produces an image of the coils 340 cm away from the mirror. How far away are the coils from the mirror? cm away. What is the absolute value of the magnification of the heater element? Magnification = . Note that its large magnitude helps spread out the reflected energy. Can you see why that is useful for this device?

An experimenter at an optical research company finds it difficult to experiment, as the measuring device that determines the angle φ inside the mirror always interferes with light progression. In this case, it would be good to measure the angle δ outside the mirror. So how does β, the relative angle of reflection with respect to the angle of incidence, relate to the angle δ?

Optical fibers are a central part of technology for rapid information transmission with a high level of efficiency. A new type of optical fiber is crafted with an index of refraction n = less than that of water in the vicinity of room temperature, where NH2O fact that n < nH20 potentially problematic if the fiber should be submerged? 1.25 = 1.33. Why is the O A. Total internal reflection would be frustrated, allowing loss of signal. O B. Internal reflection would actually be enhanced, so there is no problem. C. Electric eels might have free internet access.

X rays are typically used to kill off cancer cells, however an issue some radiologists run into is the damage an x-ray may do to surrounding tissue. Because of this, accuracy plays a key role in radiation therapy. Suppose you are a medical physicist, and you have been given the task of choosing a medium with a proper index of refraction to bend the x rays such that they only hit the tumor. Suppose you know that the x ray comes in at an angle of 20 degrees with respect to the normal of your second medium. You know the tumor expands out to 5 degrees on either side of the normal what is the minimum index of refraction needed for the x ray to hit the tumor

Please refer to Figure 2 included.   A light beam shines through air and then strikes a mirror, making an angle theta = 66.3 degrees with respect to the vertical, as shown. The reflected beam then strikes a pane of glass of thickness h = 1.01 cm, located a distance y= 32.2 cm from the mirror, and then exits the glass pane at the bottom, a horizontal distance x cm away from where it struck the mirror. What is the value of x?       74.1 cm     29.6 cm     103.7 cm     44.5 cm

Please refer to Figure 2. A light beam shines through air and then strikes a mirror, making an angle theta = 48.4 degrees with respect to the vertical, as shown. The reflected beam then strikes a pane of glass of thickness h = 1.97 cm, located a distance y= 40.0 cm from the mirror, and then exits the glass pane at the bottom, a horizontal distance x cm away from where it struck the mirror. What is the value of x?     64.7 cm   27.7 cm   46.2 cm   13.9 cm