A dentist mirror (concave) has a radius of curvature of 3 cm. How far must it be
placed from a small dental cavity to give virtual image of cavity that is magnified 5
times? (class 10 physics)
Answers
Answered by
450
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
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
suhanisharma:
sorry but your answer is wrong
you forgot that m+-v/u in mirrors
*m=-v/u
anyway, thanks for taking the effort to answer
thanks for editing your answer
yeah
ya, sorry for the little mistake :)
it's fine:)
Answered by
223
Given data:
Radius of curvature = r = -3 [because of concave mirror]
Then,
Focal Length = f = r/2 = 3/2 = -1.5
It is magnified(m) by 5 times⇒ -v/u = 5 then v = -5u
We know,
1/v+1/u = 1/f
Then put values,
⇒1/(-5u) + 1/u = 1/(-1.5)
⇒u=-1.2 cm
∴ Mirror placed at the distance of 1.2cm.
Radius of curvature = r = -3 [because of concave mirror]
Then,
Focal Length = f = r/2 = 3/2 = -1.5
It is magnified(m) by 5 times⇒ -v/u = 5 then v = -5u
We know,
1/v+1/u = 1/f
Then put values,
⇒1/(-5u) + 1/u = 1/(-1.5)
⇒u=-1.2 cm
∴ Mirror placed at the distance of 1.2cm.
Is my ans is right?
yup
i just got it on solving on my own
Great!
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