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

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


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.

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

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?

                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 = ?

As we know the formula for minima

                d sinq = (m +1/2) x l

                for 1st 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 102 x 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

                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 = ?
                d sinq = nl
                l = dsinq/n
                    = 4×10-6 sin(15)/2
                l = 518nm

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