Question

An oral surgeon wishes to inspect a patient's tooth with a magnifying mirror. He places the mirror 1.25 cm behind the tooth. This results in an upright, virtual image of the tooth that is 7.00 cm behind the mirror. 

(a) What is the mirror's radius of curvature (in cm)? 

(b) What magnification describes the image described in this passage? 

A ray of light travels through air until it strikes the interface between the air and another medium. The incident ray makes an angle of θ1 = 49.0° with the normal, as shown in the figure below. Upon passage into the second medium, the ray is refracted, emerging from the interface at an angle θ2 with respect to the normal. 

(a) Suppose that the second medium is ice. What is the angle of refraction, θ2 (in degrees)? 

(b) Suppose that the second medium is flint glass. What is the angle of refraction, θ2, in this case (in degrees)? 

(c) Finally, suppose that the second medium is glycerine. What is the angle of refraction, θ2, in this case (in degrees)? (Enter your answer to at least one decimal place.) 

Answer 1

a) The image is virtual and upright, so the image distance is d= -7.0 cm. The object is placed 1.25cm behind the tooth, so the object distance is de = 1.25 cm The mirror equation is, 1 1 1 5 do a Here, d is the object distance, d, is the image distance from the mirror, and f is the focal length of the mirror. -+- - +- Solve the equation for the focal length of the mirror. 1-11 f do di d; + do dod do di d; + do (1.25 cm) (-7.0 cm -7.0 cm+1.25 cm = 1.52 cm

The focal length of the mirror is, Here Ris the radius of the mirror. Solve the equation for the radius of curvature. R=2f = 2(1.52 cm) = 3.04 cm b) The magnification produced by the mirror is, Here d. is the object distance, and d is the image distance from the mirror. So the magnification is. ME -7.0 cm 1.25 am =5.6 As the value of magnification is greater than one and is positive, so the image formed will be upright and magnified.