Test: Mirror Formulae
15 Questions MCQ Test Science & Technology for UPSC CSE | Test: Mirror Formulae
When a mirror is rotated through an angle θ, the reflected ray from it turns through an angle of
Consider a plane-mirror and a fixed incident ray of light reflecting from the surface at an angle θi. Before the mirror has rotated, the angle of incidence is θ as is the angle of reflection. If the mirror is rotated through an angle φ the normal is rotated by an angle &phi and thus the angle of incidence increases to θ+φ. Therefore, the angle of reflection must also increase by φ to θ+φ The difference between the final angle of reflection and the initial angle of reflection and is 2φ.
Hence, For a fixed incident ray, the angle of the reflection is twice the angle through which the mirror has rotated: thetaf= 2φ
An object is placed 60 cm in front of a concave mirror. The real image formed by the mirror is located 30 cm in front of the mirror. What is the object’s magnification ?
u = -60 cm, v = -30 cm
m = – v/u = – (-60)/-30 = -0.5
A concave mirror of focal length ‘f’ produces an image ‘n’ times farther than the object. If the image is real, then the distance of the object from the mirror is:
The correct option is B.
If the magnification has a negative sign, the image formed by the concave mirror must be
We know that, if the magnification value is negative sign in the concave mirror, then the image will be real and inverted. Especially, when you come to concave mirror, the images are formed at the left of the mirror. So, it forms real and inverted.
If the magnification has a positive sign, the image formed by the concave mirror must be
It can either be a convex mirror or concave mirror. It depends upon the position of the object. u is always negative so in order to get magnification positive v must be negative. If the object is kept between pole and the focus of the mirror, then it will form a virtual image behind the mirror and v will be +.
According to the new cartesian sign convention,
When the image is of double size as the object, the magnification is
Magnification is the ratio of height of the image to the height of the object. If the image size is double of object size, then the magnification is equal to two.
If an object is 14 cm in front of convex mirror and image distance is 5.8 cm then it's focal length is
1/f = 1/v - 1/u
1/ f = 1/ 5.8 - 1/14
F = -4. 1 cm
The mirror formula is valid for
When the image is of the same size as the object, the magnification is
Which one is correct as the new cartesian sign conventions ?
Object is always kept on the left side of the mirror therefore its distance is always negative when taken from left of pole.
An object is placed on the principal axis of a spherical mirror. The object distance u is:
The unit of linear magnification is
Magnification is thr ratio of size of image to size of object and since it is a ratio of tel similar quantities hence it is unitless.
An object 2 cm high is placed at a distance of 15 cm from a concave mirror which produces an inverted image 4 cm high. Find the position of the image.
f=-10cm,u=-15cm, 1/v=1/u-1/f
[-15-(-10)]/150=-1/30,so, v=-30 cm. So, the image is formed in the same side of the object at a distance of 30 cm.
The magnification of a convex mirror is always:
The image formed by a convex mirror is always smaller in size than the object; hence its magnification is less than 1.
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