IEMDaily - Video Lecture Notes

Definition

Spherometer

A spherometer is an instrument for the precise measurement of the radius of a sphere. It is generally used for determining the radius of curvature of convex or concave mirrors and lenses. It can also be used to measure the thickness of a microscope slide or the depth of depression in a slide. The usual form consists of a fine screw moving in a nut carried on the centre of a small three-legged table or frame; the feet forming the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres, so that each rests on a point. If the screw has two turns of the thread to the millimetre the head is usually divided into 50 equal parts, so that differences of 0.01 millimetre may be measured without using a vernier. The radius of a spherometer is given by:

R = \frac{h}{2} + \frac{a^{2}}{6 h}

where:

h :

saggital (length along linear scale)

a :

length between two legs of spherometer
Methodology:

Place the spherometer on a flat surface and gently wind the screw downwards until it just touches the glass, as shown by one further division on the dial causing a just perceptible wobble.
The dial reading at this point could be noted, or alternatively the index may be circumferentially adjusted to zero by loosening the screw securing it to the table.
The instrument is then transferred to the lens or mirror to be measured, and the micrometer screw raised or lowered until all four points are just in contact with the glass.
The dial is then read for a second time, allowing the difference between the plane and curved settings to be found.
This procedure should be repeated in several orientations across the lens or mirror: a satisfactorily spherical shape would be proved by no change in the reading.

Definition

Parallax Method of Measurement

Astronomers use an effect called parallax to measure distances to nearby stars. Parallax is the apparent displacement of an object because of a change in the observer's point of view.
To measure the distance

D

of a far away planet

S

by the parallax method, We observe it from two different positions (observatories)

A

and

B

on the Earth, separated by distance

A B = b

at the same time as shown in the given figure.
We measure the angle between the two directions along which the planet is viewed at these two points. The

∠ A S B

in the figure represented by symbol

θ

is called the parallax angle or parallactic angle.As the planet is very far away,

b D << 1

and therefore,

θ

is very small.
Then we approximately take

A B

as an arc of length

b

of a circle with center at

S

and the distance

D

as the radius

A S = B S

so that

A B = b = D θ

where

θ

is in radians.

D = \frac{b}{θ}

..... (1)
Having determined

D

, we can employ a similar method to determine the size or angular diameter of the planet. If

d

is the diameter of the planet and

α

the angular size of the planet (the angle subtended by

d

at the earth),
We have

α = \frac{D}{d}

..... (2)
The angle

α

can be measured from the same location on the earth. It is the angle between the two directions when two diametrically opposite points of the planet are viewed through the telescope. Since

D

is known, the diameter

d

of the planet can be determined using

E q u a t i o n (2)

Definition

Methods of measurement of very small distances

A simple method for estimating the molecular size of oleic acid is as follows:
Oleic acid is a soapy liquid with large molecular size of the order of

10^{- 9} m

. A mono-molecular layer of oleic acid is first formed on water surface.

1 c m^{3}

of oleic acid is dissolved in alcohol to make a solution of

20

c m^{3}

1 c m^{3}

of this solution is then taken and diluted to

20 c m^{3}

, using alcohol. So, the concentration of the solution is equal to

\frac{1}{20 \times 20}

cc of oleic acid per cc of solution. Next, some lycopodium powder is lightly sprinkled on the surface of water in a large trough and one drop of this solution is put in the water. The oleic drop spreads into a thin large and roughly circular film of molecular thickness on water surface. Then, the diameter of the thin film is quickly measured to get its area

A

. Suppose we have dropped

n

drops in the water. Initially, we determine the approximate volume of each drop (

V

cc).
Volume of

n

drops of solution =

n V

cc
Amount of Oleic acid in this solution =

n V

(

\frac{1}{20 \times 20}

) cc
This solution of oleic acid spreads very fast on the surface of water and forms a very thin layer of thickness

t

. If this spreads to form a film of area

A

c m^{2}

. Then the thickness of the film:

t = \frac{V o l u m e o f t h e f i l m}{A r e a o f t h e f i l m}

or,

t = \frac{n V}{20 \times 20 A}

cm
If we assume that the film has mono-molecular thickness, then this becomes the size or diameter of a molecule of oleic acid. The value of this thickness comes out to be of the order of

109 m

Example

Problem on measurement

Problem:
The dimensional formula for a physical quantity

X

M^{- 1} L^{3} T^{- 2} .

The errors in measuring the quantities

M, L

and

T

respectively are 2%, 3%, and 4%. The maximum percentage error that occurs in measuring the quantity

X

is :
Solution:

[X] = [M^{- 1} L^{3} T^{- 2}]

\Rightarrow E r r o r = (\frac{△ M}{M} + 3 \frac{△ L}{L} + 2 \frac{△ T}{T})

\Rightarrow 2 + 3 \times (3) + 2 \times (4)

\Rightarrow

19 %

Problem:

A wire has a mass

(0.3 \pm 0.003)

g, radius

(0.5 \pm 0.005)

mm and length

(6 \pm 0.06)

cm. The maximum percentage error in the measurement of its density is :

Solution:

Formula for density is

d = m / V = m / A h = m / π r^{2} l

Δ d / d = Δ m / m + 2 Δ r / r + Δ l / l

Δ d = 0.01 + 2 \times 0.01 + 0.01

Δ d / d = 0.04

Therefore, percentage error is

Δ d / d \times 100 = 4

Example

Simple problem to find distance in light year

Example: Calculate the distance traveled by light in space in 5 light years?
Solution:
Distance traveled in 1 light year

= 9.467 \times 10^{15} m

Distance traveled in 5 light year

= 5 \times 9.467 \times 10^{15} m

= 47.34 \times 10^{15} m

Definition

Other common units of time

Smaller units of time: (1) Millisecond (2) Microsecond (3) Shake (4) Nanosecond
Bigger Units of Time: (1) Minute (2) Hour (3) Day (4) Month (5) Lunar Month (6) Year (7) Leap Year (8) Decade (9) Century (10) Millenium

Example

Some Other Common Units of Time

Some other common units of time include :
1. Second (s)
2. Millisecond (ms)
3. Minute (min)
4. Hour (h)
5. Day (d)

Formula

Measurement of area by unit square

What is the area of the shape? Each square in the grid is $1 \times 1$ , $1$ meter square.

Definition

Volume

The amount of space that a substance or object occupies, or that is enclosed within a container.
Volume has units

m^{3}

, litre and cc.

Example

Calculate time period

Problem:
A common hydrometer has

1

and

0.8

specific gravity marks

4

cm apart. Calculate the time period of vertical oscillations when it floats in water. Neglect resistance of water.
Solution:
lets say

l

is the length of the hydrometer when it is dipped in water and

A

is the cross section area of the hydrometer, then

l + 4

cm length of the hydrometer will be dipped when it is placed in liquid of specific gravity 0.8. So we have
weight of hydrometer=water displaced = liquid (0.8 s.g.) displaced