Examples on Magnification Formula
Example 1: What is the magnification produced if the image distance is 6 cm and the object is located at 12 cm in case of concave mirror?
Solution:
As we know the magnification can be calculated using the following formula;
Given, v= -6cm and u= -12cm the signs are given using sign convention.
m = -0.5
Hence, there is a decrease by a factor of 0.5.
Example 2: What is the image distance in case of convex mirror if the object is placed at 12cm? Determine it if the height of the image if 4 cm and height of the object is 2 cm.
Solution:
As we know the magnification can be calculated using the following formulas;
m and also m
Given, height of image h’ = 4cm, height of object{h}= 2cm and u= -12cm the signs are given using sign convention.
m = +2
Hence, there is an increase by a factor of 2.
Putting m= 2 and u=-12cm we get
v= 24cm
Hence, the image distance is 24cm.
Example 3: What is the magnification produced if the image distance is 12 cm and the object is located at 6 cm in case of convex mirror?
Solution:
As we know the magnification can be calculated using the following formula;
Given, v= 12cm and u= -6cm the signs are given using sign convention.
m = 2
Hence, the magnification is 2.
Example 4: What is the increase or decrease in the magnification if the object is located at 7 cm in front of a concave mirror and the image is formed at 14 cm?
Solution:
As we know the magnification can be calculated using the following formula;
m=
Given, v= -14 cm and u= -7 cm the signs are given using sign convention.
m=
m=-2
Hence, there is an increase by a factor of 0.5.
Example 5: What is the magnification if the object height is 6 cm and the image height is 24 cm below the principal axis?
Solution:
As we know the magnification can be calculated using the following formula;
Given, height of image h’ = -24 cm, height of object{h}= 6cm the signs are given using sign convention.
m=-4
Hence, the magnification is (-4).
Example 6: What is the magnification if the object height is 6 cm and the image height is 18 cm above the principal axis?
Solution:
As we know the magnification can be calculated using the following formula;
Given, height of image h’ = 18cm, height of object{h}= 6cm the signs are given using sign convention.
m=+3
Hence, the magnification is 3.
Example 7: What is the image distance in case of concave mirror if the object distance is 8 cm? It is given that the focal length of the mirror is 4 cm.
Solution:
As we know from mirror formula,
Where u= object distance= -8cm
v= image distance=?
f= focal length of mirror= -4cm
Putting values we get
v= -8 cm
Hence, the object is located 8 cm in front of the mirror.
Example 8: What is the image distance in case of convex mirror if the object distance is 10 cm? It is given that the focal length of the mirror is 10 cm.
Solution:
As we know from mirror formula,
Where u= object distance= -10cm
v= image distance=?
f= focal length of mirror= +10cm
Putting values we get
v= 5 cm
Hence, the image is located 5 cm behind the mirror.
Example 9: What is the image distance in case of concave mirror if the object distance is 11 cm? It is given that the focal length of the mirror is 11 cm.
Solution:
As we know from mirror formula,
Where u = object distance= -11cm
v= image distance=?
f= focal length of mirror= -11cm
Putting values we get
v= infinity
Hence, the image will be formed at infinity.
Mirror Formula and Magnification
The light gets reflected or refracted from the surfaces or medium. Any surface which is polished or in other words shiny always acts like a mirror be it of any kind. The observation of light bouncing off or getting back from the surfaces is termed as reflection. The light after the case of reflection travels or follows in the same medium from where the ray was incident on the surface. This phenomenon of reflection does not intend to change the velocity of light it only reverses the direction of light incident on it. This can be observed on any surface which is rough or smooth. The path of the reflected ray will depend upon the extent of smoothness of the surface, in the case of a smooth surface the reflected ray emerges with the same angle as of incidence and in the latter case suffers irregular reflection and so the reflected ray doesn’t emerge same as that of incidence angle. On the other view, light can also change its speed when traveling from one medium to another which is known as refraction of light.