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-rw-r--r--3875/CH4/EX4.14/4_13.txt1
-rw-r--r--3875/CH4/EX4.14/Ex4_13.sce13
-rw-r--r--3875/CH4/EX4.16/4_16.txt2
-rw-r--r--3875/CH4/EX4.16/Ex4_16.sce5
-rw-r--r--3875/CH4/EX4.17/Ex4_17.sce9
-rw-r--r--3875/CH4/EX4.20/Ex4_20.sce8
-rw-r--r--3875/CH4/EX4.21/Ex4_21.sce5
-rw-r--r--3875/CH4/EX4.4/Ex4_4.sce4
8 files changed, 19 insertions, 28 deletions
diff --git a/3875/CH4/EX4.14/4_13.txt b/3875/CH4/EX4.14/4_13.txt
deleted file mode 100644
index 55b42f247..000000000
--- a/3875/CH4/EX4.14/4_13.txt
+++ /dev/null
@@ -1 +0,0 @@
-The refractive index of oil is = 1.375 \ No newline at end of file
diff --git a/3875/CH4/EX4.14/Ex4_13.sce b/3875/CH4/EX4.14/Ex4_13.sce
deleted file mode 100644
index 697284dfd..000000000
--- a/3875/CH4/EX4.14/Ex4_13.sce
+++ /dev/null
@@ -1,13 +0,0 @@
-clc;
-clear;
-vol=0.2 //volume of a drop of oil in cubic centimeter
-area=100*100 //area in cm^2
-lambda=5.5*10^-5 //wavelength in m
-r=0 //angle of incidence in degree
-n=1 //number of dark band
-
-//calculation
-d=vol/area //thickness of the film of oil in cm
-myu=(n*lambda)/(2*d*cosd(0))
-
-mprintf("The refractive index of oil is = %1.3f",myu)
diff --git a/3875/CH4/EX4.16/4_16.txt b/3875/CH4/EX4.16/4_16.txt
index 31d5d83cd..e8d5b6826 100644
--- a/3875/CH4/EX4.16/4_16.txt
+++ b/3875/CH4/EX4.16/4_16.txt
@@ -1 +1 @@
- The distance at which the 10th fringe will be obtained from the edge of the wedge is =2.85e-08 m \ No newline at end of file
+ The distance at which the 10th fringe will be obtained from the edge of the wedge is = 2.85e-04 m \ No newline at end of file
diff --git a/3875/CH4/EX4.16/Ex4_16.sce b/3875/CH4/EX4.16/Ex4_16.sce
index fca7af854..e4882cfef 100644
--- a/3875/CH4/EX4.16/Ex4_16.sce
+++ b/3875/CH4/EX4.16/Ex4_16.sce
@@ -5,7 +5,6 @@ lambda=6000*10^-10 //wavelength in m
n=10 //the fringe observed
//calculation
-x=((2*n-1)*lambda)/4*alpha
+x=((2*n-1)*lambda)/(4*alpha)
-mprintf("The distance at which the 10th fringe will be obtained from the edge of the wedge is =%1.2e m",x)
-//The answer provided in the textbook is wrong.
+mprintf("The distance at which the 10th fringe will be obtained from the edge of the wedge is = %1.2e m",x)
diff --git a/3875/CH4/EX4.17/Ex4_17.sce b/3875/CH4/EX4.17/Ex4_17.sce
index df7f1f6e7..65446a8b8 100644
--- a/3875/CH4/EX4.17/Ex4_17.sce
+++ b/3875/CH4/EX4.17/Ex4_17.sce
@@ -1,13 +1,14 @@
clc;
clear;
-f=4 //focal length of lens in m
+f=400 //focal length of lens in cm
myu=1.50 //refractive index
-lambda=5460*10^-10 //wavelength in m
+lambda=5460*10^-6 //wavelength in cm
n=5 //fifth bright ring
//calculation
R=(myu-1)*2*f //radius of curvature in cm
-D5=sqrt(2*(2*n-1)*lambda*R)
-mprintf("The diameter of the fifth bright ring is = %1.5f cm",D5)
+D5=sqrt(2*((2*n)-1)*lambda*R)
+
+mprintf("The diameter of the fifth bright ring is = %1.2f cm",D5)
//The answer provided in the textbook is wrong.
diff --git a/3875/CH4/EX4.20/Ex4_20.sce b/3875/CH4/EX4.20/Ex4_20.sce
index d7db69871..5d462a526 100644
--- a/3875/CH4/EX4.20/Ex4_20.sce
+++ b/3875/CH4/EX4.20/Ex4_20.sce
@@ -3,8 +3,12 @@ clear;
R=100 //radius of curvature in cm
D_5=0.3 //diameter of the 5th dark ring in cm
D_25=0.8 //diameter of the 25th dark ring in cm
-p=20 //difference in no of fringes
+n5=5 //fifth dark ring
+n25=25 //twenty fifth ring
//calculation
-lambda=(D_25^2-D_5^2)/(4*p*R)
+p = n25 - n5 //difference in no of fringes
+lambda=((D_25^2)-(D_5^2))/(4*p*R)
+
mprintf("The wavelength of light used is = %1.2e cm",lambda)
+//The answer provided in the textbook is wrong.
diff --git a/3875/CH4/EX4.21/Ex4_21.sce b/3875/CH4/EX4.21/Ex4_21.sce
index e57549561..c0b88ce7f 100644
--- a/3875/CH4/EX4.21/Ex4_21.sce
+++ b/3875/CH4/EX4.21/Ex4_21.sce
@@ -1,9 +1,10 @@
clc;
clear;
-X2_minus_X1=0.05896*10^-3 //displacement of mirror in m
+disp_X=0.05896*10^-3 //displacement of mirror in m
n=200 //no of fringes
//calculation
-lambda=(2*X2_minus_X1)/n
+lambda=(2*disp_X)/n
+
mprintf("The wavelength of the light is = %1.3e m",lambda)
//The answer provided in the textbook is wrong.
diff --git a/3875/CH4/EX4.4/Ex4_4.sce b/3875/CH4/EX4.4/Ex4_4.sce
index 2c723e8f2..392885aab 100644
--- a/3875/CH4/EX4.4/Ex4_4.sce
+++ b/3875/CH4/EX4.4/Ex4_4.sce
@@ -6,8 +6,8 @@ d1=4.05*10^-3 //distance between two images of the slit in m in first case
d2=2.90*10^-3 //distance between two images of the slit in m in second case
//calculation
-d=sqrt(d1*d2)
-disp(d)
+d=sqrt(d1*d2) //distance between the slits in m
Beta=(lambda*D)/d
mprintf("The distance between interference fringes is %1.2e m",Beta)
+//The answer provided in the textbook is wrong.