Modified the code

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.