A dentist has a small mirror of focal length 1.6 cm. He observes the cavity in the tooth of a patient by holding the mirror at a distance of 8 mm from the cavity. The magnification is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
A dentist uses a concave mirror and holds it closer to the tooth than its focal length, so that it forms an enlarged, erect, virtual image. Use the mirror formula with the Cartesian sign convention (distances measured from the pole, positive in the direction of the incoming light).
\( f = -1.6\,\mathrm{cm} \)
\( u = -8\,\mathrm{mm} = -0.8\,\mathrm{cm} \)
\( \displaystyle \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{1.6} + \frac{1}{0.8} = \frac{1}{1.6} \)
\( \displaystyle v = +1.6\,\mathrm{cm} \)
The positive \( v \) shows that the image is behind the mirror (virtual).
\( \displaystyle m = -\frac{v}{u} = -\frac{1.6}{-0.8} = 2 \)
The image is erect and twice the size of the cavity.