Question

A dentist uses a concave mirror to examine a tooth that is 1.00 cm in front of the mirror. The image of the tooth forms 10.0 cm behind the mirror.What is the mirror’s radius of curvature?What is the magnification of the image?

Answer 1

Here ,

object distance (u)= -1.00 cm

image distance(v)= -10.0 cm

focal length = -f

radius of curvature =r

Using concave mirror formula

`\begin{gathered} (1)/(f)=(1)/(u)+(1)/(v); \\ \therefore(1)/(-f)=(1)/(-1)+(1)/(-10); \\ (1)/(f)=\text{ }(11)/(10); \\ \therefore f=\text{ }(10)/(11); \end{gathered}`

Again we have

r= 2f

`\begin{gathered} r=\text{ 2f } \\ \therefore r=\text{ 2}*(10)/(11)\text{ =1.818 cm = 1.82cm\lparen Approx\rparen} \end{gathered}`

answer is 1.82 cm ( negative in sign)

Now magnification is

`magnification\text{ = }(v)/(u)=\text{ }(10)/(1)\text{ = 10 }`

Answer is radius = 1.82cm( negative in sign) & magnification =10