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A dentist uses a small mirror that gives a magnification of 4 when it is held 0.60cm from a tooth. The radius of curvature of the mirror is
Answer
463.5k+ views
Hint: Substitute the given values in the expression for magnification and thus find the mage distance. Now recall the mirror formula for spherical mirrors and then substitute accordingly and then find the focal length. Also, we know that the radius of curvature is twice the focal length and hence we get the answer.
Formula used:
Magnification,
$m=-\dfrac{v}{u}$
Mirror formula,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Radius of curvature,
$r=2f$
Complete step-by-step answer:
For spherical mirrors, magnification is the increase in image size produced with respect to the object size. That is, it is the ratio of height of the image to the height of the object. Magnification m produced by spherical mirrors is also equal to the ratio of image distance to the object distance. That is,
$m=-\dfrac{v}{u}$
In this case the object is kept in front of the mirror, so, by convention, we can take the object distance negative. So, object distance u is given as,
$u=-0.60cm$ ……………………. (1)
Magnification of the mirror is,
$m=4$
Substituting these values,
$4=\dfrac{-v}{-0.60}$
$v=2.4cm$ ………………………………. (2)
Now, let us recall the mirror formula for spherical mirrors given by,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\Rightarrow f=\dfrac{uv}{u+v}$
Where, f is the focal length of the mirror.
Substituting (1) and (2) we get,
$f=\dfrac{-0.60\times 2.4}{-0.6+2.4}$
$\Rightarrow f=-0.8cm$
Now we have to find the radius of curvature of the mirror. We know that the radius of curvature of a spherical mirror is twice its focal length. Therefore, radius of the given mirror is,
$r=2f=2\left( -0.8cm \right)=-1.6cm$
Hence, we find the radius of curvature of the given mirror as -1.6cm. Since the radius is found negative, by convention we know that the given mirror is concave.
Note: Remember that positive magnitude of magnification indicates virtual and erect image, while negative magnitude indicates real and inverted image. In the question we found that the mirror used by a dentist is a concave mirror. This is because a concave mirror gives the dentist a magnified reflection of the mouth, that is, the image formed will be larger, brighter and also erect making it ideal for dentist’s purpose.
Formula used:
Magnification,
$m=-\dfrac{v}{u}$
Mirror formula,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
Radius of curvature,
$r=2f$
Complete step-by-step answer:
For spherical mirrors, magnification is the increase in image size produced with respect to the object size. That is, it is the ratio of height of the image to the height of the object. Magnification m produced by spherical mirrors is also equal to the ratio of image distance to the object distance. That is,
$m=-\dfrac{v}{u}$
In this case the object is kept in front of the mirror, so, by convention, we can take the object distance negative. So, object distance u is given as,
$u=-0.60cm$ ……………………. (1)
Magnification of the mirror is,
$m=4$
Substituting these values,
$4=\dfrac{-v}{-0.60}$
$v=2.4cm$ ………………………………. (2)
Now, let us recall the mirror formula for spherical mirrors given by,
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$
$\Rightarrow f=\dfrac{uv}{u+v}$
Where, f is the focal length of the mirror.
Substituting (1) and (2) we get,
$f=\dfrac{-0.60\times 2.4}{-0.6+2.4}$
$\Rightarrow f=-0.8cm$
Now we have to find the radius of curvature of the mirror. We know that the radius of curvature of a spherical mirror is twice its focal length. Therefore, radius of the given mirror is,
$r=2f=2\left( -0.8cm \right)=-1.6cm$
Hence, we find the radius of curvature of the given mirror as -1.6cm. Since the radius is found negative, by convention we know that the given mirror is concave.
Note: Remember that positive magnitude of magnification indicates virtual and erect image, while negative magnitude indicates real and inverted image. In the question we found that the mirror used by a dentist is a concave mirror. This is because a concave mirror gives the dentist a magnified reflection of the mouth, that is, the image formed will be larger, brighter and also erect making it ideal for dentist’s purpose.
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