Physics Times 07.2019

^

4.Sol:

Mg N Ma  v

16

Mg

 N

3.Sol: As 2

1 13.6

&

eV

E E

n









4 2

4 1

1 1

16 1

/

(^15)
1
16
m
a e
E E m
E E
 
 
  
 
On solving we get
(^) m 2
1242 16
487
e 13.6 3
 nm

 

As 2
1
K
n

2
2
2
1
1 1
2 4
K
K
  as kinetic energy is proportional
(^2)
1
n
The net path difference xis
(^)  x dsindsin
sin tan
y
b
 
dy
x d
D
  
If (^)  0
/ 0.3 11 33 10^4
1000
x dy D mm

     
x interms of
4
6
33 10 11
600 10 2

 



 

So destructive interference takes place
At Point P
0.36 0.3mm 11mm
0.3mm
180 1000
x



    
(^) 39 10 mm ^4
9 3
39 10 (2 1)^4 600 10 10
2
n

      
n 7
There will be destructive interference
At Point O
0.36
3mm 0 600nm
180
x


     
600nm n 
n 1
Constructive interference
Fringe width does not depend on .
5.Sol: Case-I:
H cm 30
(^) n3/2
(^1)
30 2
/ 20cm
3
H H n

  
Case-II:
R cm 300
3.Sol:
4.Sol:
At Point P
At Point O
5.Sol: Case-I:
Case-II:
n n n n 2 1 2 1
v u R

 
2
1 3
(^132)
H 2 30 300

 
   