Question

A concave (converging) dentist's mirror, with a radius of curvature of 20 cm, is held inside a patient's mouth at a distance 5 cm from a cavity in the number 18 molar. The cavity is 0.8mm across.

a) Where will the image of the cavity appear, relative to the position of the mirror, and will it be real or virtual?

b) How big will the cavity appear in the image, and will it be erect or inverted?

Answer 1

Note - Here I am using Cartesian sign convention

a)

Using mirror equation

1/f =1/v + 1/u

here f=R/2=-20/2 =-10cm , u= -5 cm

-1/10=1/v-1/5

1/v=-1/10+1/5

1/v= (-1+2)/10

v=10 cm  

here v is positive so image is virtual.

b)We know that

m=-v/u. Here v=10cm, u=-5cm

m=10/5 =2

h' =2h =2×0.8 =1.6 mm

h' =1.6mm Answer

since m is positive so image is upright (react)