FSC 1st Year Physics Chapter 9 Exercise & Numericals Notes PDF

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In this post, I am sharing “Physical Optics” – FSC 1st Year Physics Chapter 9 Solved Exercise & Numericals Notes PDF Download for the students of Intermediate Part 1. This is the 9th chapter from the 11th Physics textbook and its name is Physical Optics. Students can download the 11th class physics chapter 9 Exercise & Numericals Waves Notes PDF on their laptop or mobile. These Physics Notes for FSC First Year are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. If you need complete FSC 1st Year Physics Notes PDF of all chapters free, go here.

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FSc Part 1 Physics Chapter 9 Physical optics Exercise & Numericals Notes PDF Download

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11th Class Physics Chapter 9 Solved Numericals Notes Pdf Download

Light of wavelength 546nm is allowed to illuminate the slits of Young’s experiment. The separation between the slits is 0.10nm and the distance of the screen from the slit where interference effects are observed is 20cm. at what angle the first minimum will fall? What will be the linear distance on the screen between adjacent maxima?

Data:
wavelength of light l = 546nm = 546 x 10^-9 separation between slits = d = 0.10mm
= 0.01m/1000 = 0.10 x 10³

                Distance of screen from the slits = L = 20 cm = 20 x 10 ²
                Angle for first minimum = q = ?
                fringes spacing = Dy = ?
calculation:

As we know the formula for minima

d sinq = (m +1/2) x l

                for 1^st minimum m = 0
                d sinq =l/2
                sinq = l/2d
putting the value we get
                sinq = 546 x10^-9/ 2 x 0.10 x 10^-3                sinq = 0.00273
                     q = sin^-1(0.00273)
                     q = 0.16⁰
using the formula for fringe spacing
                   Dy =Ll/d
by putting value we get
                    Dy= 20 x 10²x 546 x10⁹/0.10 x 10^-3
                   Dy= 1.1mm (Ans)

A light is incident normally on a grating which has 2500 lines per centimetre. Compute the wavelength of a spectral line for which the deviation in second order is 15.0⁰

Data
                number of line on grating = N = 2500lines/cm
                d = 1/N = 1/2500 = 4 x 10^-4 lines/m
                For second-order = n = 2
                angle of deviation = q = 15
                wavelength of light = l = ?
Calculation:
                d sinq = nl
                l = dsinq/n
                    = 4×10^-6 sin(15)/2
                l = 518nm

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