Optics Concept Page - 9

Definition

Minimum distance between object and its real image for a concave mirror

The minimum distance between an object and its real image in case of a concave mirror is zero when the object is placed at the centre of curvature. In this case, the image also is formed at the centre of curvature.

Example

Image of an object of finite size along the principal axis for a spherical mirror

An object 5 cm$5cm$ tall is placed 1 m$1m$ from a concave spherical mirror, which has a radius of curvature of 20 cm$20cm$. What is the size of the image?Here
u=−1 m$u=-1m$ and f=−10 cm$f=-10cm$
Using 1u+1v=1f${\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}$, we get:
1−100+1v=1−20${\displaystyle \frac{1}{-100}}+{\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{-20}}$
or, 1v=−4100${\displaystyle \frac{1}{v}}={\displaystyle \frac{-4}{100}}$
or, v=−1004$v={\displaystyle \frac{-100}{4}}$
M=−vu=h2h1$M={\displaystyle \frac{-v}{u}}={\displaystyle \frac{h_2}{h_1}}$
or, −1004100=h2h1$-{\displaystyle \frac{{\displaystyle \frac{100}{4}}}{100}}={\displaystyle \frac{h_2}{h_1}}$
h1=−54=−1.25 cm$h_1={\displaystyle \frac{-5}{4}}=-1.25cm$

Definition

Magnification

Magnification is defined as the ration of height of image to height of object.
m=IO=−vu$m={\displaystyle \frac{I}{O}}=-{\displaystyle \frac{v}{u}}$
Note:
Sign convention must be followed while using formula for magnification. Hence, it can be positive or negative.
|m|>1⟹$|m|>1⟹$ image is magnified.
|m|=1⟹$|m|=1⟹$ image is same size as object.
|m|<1⟹$|m|<1⟹$ image is diminished.

Example

Problem on mirror equation

Problem:
If an object is placed 10 cm in front of a convex mirror of focal length 20 cm, then distance of the image from the mirror is:
Solution:
Given;focal length f=20cm$f=20cm$
          Object distance u=−10cm$u=-10cm$
Uing mirror formula.
1v+1u=1f$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
1v=1f−1u$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
      =120−1−10$=\frac{1}{20}-\frac{1}{-10}$         (∵u=−10$\because u=-10$)
      =120+110$=\frac{1}{20}+\frac{1}{10}$
      =320$=\frac{3}{20}$
Image distance⇒v=203$\Rightarrow v=\frac{20}{3}$cm

Example

Problems involving the mirror equation

If an object is placed 10 cm$10cm$ in front of a convex mirror of focal length 20 cm$20cm$, what will be the distance of the image from the mirror?Given; f=20 cm$f=20cm$, u=−10 cm$u=-10cm$
Using mirror formula, 
1v+1u=1f${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
1v=1f−1u${\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}-{\displaystyle \frac{1}{u}}$=120−1−10$={\displaystyle \frac{1}{20}}-{\displaystyle \frac{1}{-10}}$         (∵u=−10$\because u=-10$)
=120+110$={\displaystyle \frac{1}{20}}+{\displaystyle \frac{1}{10}}$=320$={\displaystyle \frac{3}{20}}$
⇒v=203 cm$\Rightarrow v={\displaystyle \frac{20}{3}}cm$

Example

Speed of object and speed of image

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R=2 m$R=2m$. If the jogger is running at a speed of 5 ms−1$5m{s}^{-1}$, how fast the image of the jogger appear to move when the jogger is 39 m$39m$ away?
Solution:
From the mirror equation, we get:
v=ufu−f$v={\displaystyle \frac{uf}{u-f}}$
For convex mirror, since R=2 m,f=1 m$R=2m,f=1m$, then for u=39 m$u=39m$, v=−39×1−39−1$v={\displaystyle \frac{-39\times 1}{-39-1}}$
Since the jogger moves at a constant speed of 5 ms−1$5m{s}^{-1}$, after 1 s$1s$, the position of the image v$v$ (for u=39−5=34$u=39-5=34$) is 3435 m${\displaystyle \frac{34}{35}}m$.
The shift in the position of image in 1 s$1s$ is
3940−3435=1280 m${\displaystyle \frac{39}{40}}-{\displaystyle \frac{34}{35}}={\displaystyle \frac{1}{280}}m$
Therefore, the average speed of the image when the jogger is between 39 m$39m$ and 34 m$34m$ from the mirror, is 1280 ms−1${\displaystyle \frac{1}{280}}m{s}^{-1}$.

Definition

Refraction of light

Definition:
The direction of propagation of an obliquely incident ray of light that enters the other medium, changes at the interference at the interface of the two media. This phenomenon is called as refraction of light.

Refraction is the change in direction of wave propagation due to a change in its transmission medium.

The phenomenon is explained by the conservation of energy and the conservation of momentum. Due to change of medium, the phase velocity of the wave is changed but its frequency remains constant.

Definition

Optical density of medium

Definition:
The optical density of a material is a logarithmic ratio of the falling radiation to the transmitted radiation through a material. It is also referred as a fraction of absorbed radiation at a particular wavelength.

The speed of light depends on the characteristics of the medium on which it is incident; the optical density of the medium influences the speed of light.
Optical density of the medium refers to the sluggish tendency of the atoms of a material to retain the energy absorbed from the electromagnetic wave in the form of vibrating electrons before being reemitted as an electromagnetic disturbance.

The refractive index of the material is an indicator of its optical density. Spectrometer is used to measure the optical density of a material.

Definition

Rarer and Denser Medium

When light refracts from one medium to another, one of the medium is termed as rarer and other is termed as denser. Hence rarer and denser medium is a relative term. The medium in which speed of light is more is termed as a rarer medium and the medium in which the speed of light is less is termed as a denser medium.
When light refracts from rarer to denser medium it bends towards the normal as shown in the figure above.

Definition

Refractive index

Refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in air.
μ or n=cv$\mu \text{ or }n={\displaystyle \frac{c}{v}}$
Note: μ≥1$\mu \ge 1$
          It is usually found using Snell's laws of refraction.

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