Wednesday, September 28, 2022

10. Light - Reflection and Refraction | Class 10 CBSE | Web Notes | Part 4: Sign Convention, Mirror Formula & Magnification

10. LIGHT – REFLECTION AND REFRACTION

Sign Convention for Reflection by Spherical Mirrors


New Cartesian Sign Convention:


In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x-axis (X’X) of the coordinate system.


The New Cartesian Sign Convention for spherical mirrors


The conventions are as follows:

a) The object is always placed to the left of the mirror. i.e., light from the object falls on the mirror from the left-hand side.

b) All distances parallel to the principal axis are measured from the pole of the mirror.

c) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along – x-axis) are taken as negative.

d) Distances measured perpendicular to and above the principal axis (along + y-axis) are taken as positive.

e) Distances measured perpendicular to and below the principal axis (along – y-axis) are taken as negative.


Sign conventions are applied to obtain the mirror formula and solve related numerical problems.


Mirror Formula and Magnification

In a spherical mirror, the distance of the object from its pole is called the object distance (u).


The distance of the image from the pole of the mirror is called the image distance (v).


The distance of the principal focus from the pole is called the focal length (f).


This formula is valid in all situations for all spherical mirrors for all positions of the object.


Magnification (m)


It is the enlargement of the image formed by a spherical mirror, relative to the size of the object.


It is the ratio of the height of the image (h′) to the height of the object (h).


Magnification is also related to the object distance (u) and image distance (v). It can be expressed as:


The height of the object is taken to be positive as the object is placed above the principal axis.


The height of the image is taken as positive for virtual images and negative for real images.


A negative sign in the value of the magnification indicates that the image is real. A positive sign indicates that the image is virtual.


Problem: A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image.


Solution

Radius of curvature, R = + 3.00 m

Object-distance, u = – 5.00 m
Image-distance, v= ?

Height of the image, h′= ?

Focal length, f = R/2 = + 300 m/2 = +1.50 m (as the principal focus of a convex mirror is behind the mirror)

The image is 1.15 m at the back of the mirror.


The image is virtual, erect & smaller by a factor of 0.23.


Problem: An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.


Solution

Object-size, h = + 4.0 cm

Object-distance, u = – 25.0 cm

Focal length, f = –15.0 cm

Image-distance, v= ?

Image-size, h′= ?


v = – 37.5 cm


The screen should be placed at 37.5 cm in front of the mirror. The image is real.

Height of the image, h′ = – 6.0 cm

The image is inverted and enlarged.


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