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A dentist's mirror has a radius of curvature of 3cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?
Most Upvoted Answer
A dentist's mirror has a radius of curvature of 3cm. How far must it b...
Given ;
R = -3 cm
f = -1.5 cm

m = -v/u
5 = -v/u
5u = -v

Using mirror formula;

1/f = (1/v)+(1/u)
(-1/1.5) = (-1/5u) + (1/u)
(-10/15) = (-1+5)/5u
-2/3 = 4/5u
-10u = 12
u = -12/10
u = -1.2


Hence the mirror should be placed 1.2cm away from the cavity to produce 5 times magnification
Community Answer
A dentist's mirror has a radius of curvature of 3cm. How far must it b...
Answer:


Given:


  • Radius of curvature of the mirror = 3 cm

  • Magnification of the image = 5



To find: Distance between the mirror and the dental cavity


Explanation:


When an object is placed in front of a concave mirror, a virtual image is formed behind the mirror. The distance between the object and the mirror, and the radius of curvature of the mirror determines the position and size of the image formed.


Formula:

The formula for magnification is given by:

m = -v/u

Where,


  • m = Magnification of the image

  • u = Distance of the object from the mirror

  • v = Distance of the image from the mirror



The formula for the focal length of a concave mirror is given by:

f = R/2

Where,


  • f = Focal length of the mirror

  • R = Radius of curvature of the mirror



Solution:


Let the distance between the mirror and the dental cavity be 'x'.


From the given data, the magnification of the image is 5. Therefore,

m = v/u = 5


Using the formula for magnification, we can write:

v/u = 5

=> v = 5u


Now, using the formula for the focal length of the mirror, we can write:

f = R/2 = 3/2 cm


Using the mirror formula, we can write:

1/f = 1/u + 1/v


Substituting the values of f and v in the above equation, we get:

1/(3/2) = 1/u + 1/5u

=> 2/3 = (5+1)/5u

=> 2/3 = 6/5u

=> u = 5/9 cm


Therefore, the distance between the mirror and the dental cavity is:

x = u + f = 5/9 + 3/2 = 31/18 cm


Conclusion:


The distance between the mirror and the dental cavity is 31/18 cm.
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A dentist's mirror has a radius of curvature of 3cm. How far must it be placed from a small dental cavity to give a virtual image of the cavity that is magnified five times?
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