give any four sign conventitions for mirror
Answers
Answer:
Object is always placed to the left of mirror. All distances are measured from the pole of the mirror. Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative.
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Answer:
u is the Object distance
v is the Image distance
f is the Focal Length given by f=R2
R is the radius of curvature of the spherical mirror
The above formula is valid under all situations for all types of spherical mirrors (Concave and Convex) and for all object positions.
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Sign Conventions
New Cartesian Sign Convention
New Cartesian Sign Convention is used to avoid confusion in understanding the ray directions. Refer to the diagram for clear visualization.
For the measurement of all the distances, the optical center of the lens is considered.
When the distances are measured opposite to the direction of the incident light, they are considered to be negative.
When the distances are measured in the same direction of the incident light, they are considered to be positive.
When the heights are measured upwards and perpendicular to the principal axis, they are considered to be positive.
When the heights are measured downwards and perpendicular to the principal axis, they are considered to be negative.
Mirror Equation for concave mirror and Mirror Equation for a convex mirror
Mirror Equation for Concave and Convex Mirrors
The mirror equation 1v+1u=1f holds good for concave mirrors as well as convex mirrors.
Example of Mirror Equation
The radius of curvature of a convex mirror used for rearview on a car is 4.00 m. If the location of the bus is 6 meters from this mirror, find the position of the image formed.
Solution:
Given:
The radius of curvature (R)= +4.00 m
Object distance(u) = -6.00 m
Image distance(v) = ?
Formula used:
f=R2 1v+1u=1f
Calculation:
To calculate the Focal length of the given mirror, substitute the value of Radius of Curvature (R) in the f=R2. We get-
f=+4.00m2=+2m
Since, 1v+1u=1f we can re- arrange it as –
1v=1f−1u
On substituting the values in the above equation we get-
⇒1v=1+2.00−1(−6.00) =12.00+16.00 6+22×6=812 v=128
= 1.5 meter.
The image is 1.5 meters behind the mirror.
The mirror is a polished surface which reflects the incident light to form the image. Here reflected light will have a wavelength and many other physical properties almost the same as that of the incident light.
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