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Question

An object is placed in front of a convex mirror of radius of curvature 40cm at a distance of 10cm. Find the position and nature of the image.


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Solution

Step 1: Given data:

Radius of curvature(R) of convex mirror = +40cm

Object distance(u) = -10cm

( Note: Sign convention:

  1. All the distance to the left of the pole of a spherical mirror is negative(-) and the distances taken to the right of the pole are positive(+).
  2. Thus, the sign of the radius of curvature(R) of the convex mirror is positive(+) as it lies at the left of the pole.
  3. Object distance(u)is negative(-) as it is always considered to the left of the mirror.)

Step 2: formula used:

  1. The radius of curvature(R) and the focal length(f) of a spherical mirror are related as;

Radiusofcurvature(R)=2×Focallength(f)Or,Focallength(f)=Radiusofcurvature(R)2

2. Object distance(u), image distance(v) and focal length(f) of a mirror are related as:

1v+1u=1f

3. Magnification of a mirror(m)= -Imagedistance(v)Objectdistance(u)

(Note: Negative magnification suggests a real and inverted image whereas a positive magnification suggests a virtual and erect image.)

Step 3: Calculation of focal length(f) of the convex mirror:

f=+40cm2f=+20cm

Step 4: Calculation of position and nature of image:

The position of the image or image distance(v) will be,

1v+1-10cm=1+20cm1v-110=+1201v=120+1101v=1+2201v=320Or,v=+203cmv=+6.67cm

Now,

m=-+6.67cm-10cmm=+6671000m=0.667(Positive)Whichsuggestsavirtualanderectimage.

Hence,

The image of the truck will be at 6.67cm behind the mirror and it will be a virtual and erect image.


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