Question

It is known that the refractive index u of the material of a prism,depends on the wavelength as follows:
u=a+b/λ^2
Where a,and b are constants.Plot a graph between u and λ.

Main question : Identify the pair of variables to be used to get a straight line graph

Answer 1

When plotted against lambda, it will curve, getting large at small lambda.

Plotting y against (1/lambda^2) or plotting log y vs log lambda could produce linear (straight line) plots.

Answer 2

how do i know which pair gives straight line? @douglaswinslowcooper

Answer 3

Of course you could plot them. I chose them because they are of the form
z = m x + b, a linear equation for z in variable x

in the first suggestion z is y and x is (1/lambda^2)

in the second, z is log y and x is log(1/lambda^2) = -2 log (lambda)

Answer 4

@ganeshie8

Answer 5

\(\large \mathbb{u=a+b/\lambda^2} \)

\(\large \mathbb{u-a= b/\lambda^2} \)

take log both sides

Answer 6

\(\large \mathbb{\log (u-a)= \log(b/\lambda^2)}\)


\( \large \mathbb{\log (u-a)= \log b - \log \lambda^2}\)

\( \large \mathbb{\log (u-a)= \log b - 2\log \lambda} \)

put \(y = \log(u-a)\)
\(x = \log \lambda \)

Answer 7

\(y = \log b - 2x\)

straight line, wid slope = -2,
y-intercept = log b

Answer 8

so the new change of variables wud be :
log(u-a) and log(lambda)

Answer 9

book mein Plotting u against (1/lambda^2) likha hai :/ ill prefer to know how did we get that combination !

Answer 10

that also works right ?

Answer 11

simply put 1/lambda^2 = x
and stare at the equation

Answer 12

u=a+b/λ^2

u = a + bx

straight line

Answer 13

oh you're asking, how to get to that analytically ?

Answer 14

yes

Answer 15

let me think a bit,
logifying is the dumb method i remember...

Answer 16

okay forget the analytical method :P us wale se kaam chal jaega 8|

Answer 17

wait a sec,
logifying is good when u have an exponent funtion like : u = b/2^λ

Answer 18

but wat you're having is simply a hyperbola...
so just parameterizing x = 1/λ^2 will do i hope...

sorry cant provide anything useful :|

Answer 19

even if I have something like
\[\Huge \mu=a+\frac{b}{\lambda^{100000}}\]
i can put 1/lambda^100000000000 = x and it will still be a straight line??

Answer 20

ofcourse, if \(\mu\) and \(\lambda\) are the variables
and \(a, b\) are constant,

given equation becomes :
\(\mu = a + bx\)

Answer 21

which is a linear equation in two variables : \(\mu , x\)
produces a straight line

Answer 22

that is cool...:o

Answer 23

thanku!

Answer 24

np :)
but read a bit on logifying... im sure u wil find somthing useful on that..
ppl wont simply put x = 1/(\lambda)^100..

and convert it to straight line.. .

that doesnt help u in anyway....

Answer 25

i mean, u r not getting any benifit by simply putting x = 1/\lambda^2 or watever

the graph looks straight line, but wat u wil be acheving by that ?

you're masking the actual relatioonship of y = a + b/lambda^2

Answer 26

lol yes practically not correct :P

Answer 27

is there any quantity in optics, that is related to wavelength by :
q = k/lambda^2

if so, then i smell some use in putting simply x = 1/lambda^2 :)

Answer 28

umm no :|

Answer 29

1/lambda = f
right ?

Answer 30

no :o

Answer 31

oops !
1/T = f

Answer 32

oh that lambda,yes

Answer 33

T = time period
W = wavelength
f = frequency

Answer 34

T : seconds
W : meters
f : hertz

Answer 35

W = lambda