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Find the size, nature and position of the image formed when an object of size 1cm is placed at a distance of 15cm from a concave mirror of focal length 10cm .

Answer
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Hint: Use the mirror equation to derive the position of image. Use the magnification of a mirror formula to obtain the size of the object. Accordingly depict its nature using the sign of the image height.


Formula Used:
The mirror equation
1v+1u=1f
Where v is the image distance,u is the object distance and f is the focal length.
Also the magnification of mirrors is given by
m=vu=hiho
Where v is the image distance,u is the object distance and f is the focal length, m is the magnification, hi is the height of the image and ho is the height of object.


Complete step by step answer:
Given,
 u=15cm {Using Sign Convention}
f=10cm {Using sign convention since focal length of concave mirror is negative}
ho=1cm We can calculate v by using the following equation,
1v+1u=1f
Substituting the values of u and f in the above equation we get,
1v+115=110
1v=110+115
Taking L.C.M,
1v=2330
1v=130
Taking the reciprocal, we get
v=30cm
Thus the image is formed 30cm in front of the mirror.
Now using the equation,
m=vu=hiho
We substitute the values here,
m=(30)15=hi1cm
On solving we get,
m=2=hi1cm
Thus the image is twice the size of object.
On solving further we get that,
ho=2cm
The negative sign indicates that the image is virtual and inverted. This is also because the image is formed at the same side of the mirror with the object.
Hence finally, v=30cm , |ho|=2cm and the image is real and inverted.

Note: Do not confuse the mirror formula with the lens formula and write the magnification formula correctly. Also do not forget to use sign convention correctly as it will cause an error in the answer. Take care of the negative sign in the image height.