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Solutions : MCQ : Prism & Dispersion of Light
01. From A +  = i + e, given the prism is
equilateral,  A = 60o, also, i = e = 3A/4
 A +  = 2 e = 2  3  60 / 4
 60 +  = 90   = 30o.
(D)
02. In case of minm deviation, r1 = r2 = r
and A = 2r  A = 2  30 = 60, also it is given
that, m = 30o, hence by formula of prism
  = Sin [ ( A + m ) / 2 ] / Sin A/2
  = Sin 45 / Sin 30   = (1/2) / (1/2)
  =  2.
(A)
03.  = Sin [ ( A + m ) / 2 ] / Sin A/2
  2 = Sin [ ( 60 + m ) / 2 ] / Sin 60/2
  2 = Sin [ ( 60 + m ) / 2 ] / Sin 30
  2 = 2 Sin [ ( 60 + m ) / 2 ]
 1 /  2 = Sin [ ( 60 + m ) / 2 ] = Sin 45
 ( 60 + m ) / 2 = 45
 60 + m = 90  m = 30.
Now, i = e, hence from i + e = A + 
 2i = 60 + 30  i = 45.
(B)
04. As material of all the objects is same, hence
dispersive power will also be the same.
(D)
05. m = A,   = 2 cos A/2
1.5 = 2 cos A/2  cos A/2 = 0.75
 A/2 = cos –1 ( 0.75 ) = 41o
 A = 82o.
(C)
06. For minm deviation, A = 2r and
m = (  - 1 ) A = (  - 1 ) 2r = ( 1.5 – 1 ) 2r
m = 0.5  2r = r.
(B)
07. As the light incidents normally, hence,
i = 0, r1 = 0  A = r1 + r2  r2 = 30.
Hence for the refraction at the other face
 = Sin e / Sin r2  1.5 = Sin e / Sin 30
 Sin e = 1.5 / 2 = 0.75  e = Sin –1 ( 0.75 )
 e = 48o36/. Now A +  = i + e
 30 +  = 0 + 48o36/   = 18o36/.
(B)
08. i = 45o, r = 30o
30
 = Sin i / Sin 300
o
  = Sin 45o / Sin 30o
45
30
  = ( 1/2) / ( 1/2)
  = 2.
(A)
09. For resultant deviation zero ( qáʽààtã âwjvÂà ÎàåÂu
ÑàçÂàç Ñçmä) A/ = - (  - 1 )A / ( / - 1 )
 A/ = - ( 1.54 – 1 ) 4o / ( 1.72 – 1 )
 A/ = - 54 × 40 / 72 = 3o.
(B)
10. Again, A/ = - (  - 1 )A / ( / - 1 )
 A/ = - ( 1.54 – 1 ) 6o / ( 1.72 – 1 )
 A/ = - 54 × 60 / 72 = 9/ 2 = 4o 30/
(B)
11. Given,  = cot A/2, &
 = Sin [ ( A + m ) / 2 ] / Sin A/2
cos A / 2
sin( A +  m ) / 2
o
o
=
sin A / 2
sin A / 2
 cos A/2 = sin (A + m) / 2
 sin ( 90 – A/2 ) = sin (A + m) / 2
 90 – A/2 = A + m / 2  m = 180 – 2A
(B)
13. Angular dispersion ( §ýàç½àãu âwÕ àçq½à )
 = ( V - R ) A = ( 1.6 – 1.5 ) × 5o
  = 0.5o.
(C)
14. For equilateral  ( ytràÑC âØàsäk Ñçmä ) A = 60o and
in minm deviation position ( ¥wÞ ÂuåÂàmt âwjvÂà §ýL ¡ wÐnà
tçÞ ) i = e = 54o. Hence from
i + e = A + m  60 + m = 2 × 54
 m = 108 – 60 = 48o.
(C)
o
45
15. As  = 2,
 sin C = 1 /  = 1 / 2
o
45
 C = 45o
Hence the light ray will exit parallel to the
Second face ( ¡ mß Zà§ýàÎà â§ýʽà ÀåyÊã ymÑ §çý ÐqÎàJu âÂàªàêmî
Ñàçªàã).
àã
(A)
16. By the property of right angle prism
( yt§ýàç½à âZà³t §çý ªàä½à yç ) correct option is –
(A)
17. For TIR, Sin C = 1/,  Sin C = 1 / 1.5
 Sin C = 10 / 15 = 0.66  C = Sin -1 0.66
which lies somewhere between 300 & 450.
0
0
( §íýàÞâm§ý §ýàç½à C §ýà tàÂà 30 ¥wÞ 45 §çý rãj §ýÑãA qÊ Ñàçªàà )
Now,  = Sin i / Sin r1,  1.5 = Sin 300/Sin r1
 Sin r1 = 1 / 3 = 0.33, hence r1 must be less
0
than 300. ( r1 §ýà tàÂà 30 yç §ýt Ñàçªàà ).
Now, r1 + r2 = A,  r2 = 900 – Sin -1 ( 0.33 )
 r2 > 60 0. Hence, by above calculations it is
clear that r2 > C. Hence the ray cannot exit from
the second face ( ¡ mß Zà§ýàÎà â§ýʽà ÀåyÊã ymÑ yç âÂàªàêmî ÂàÑãA Ñàç
q४àã)
q४àã
(D)
18. As material of both the prisms are same,
hence 1 = 2. 1 : 2 = 1 : 1.
(A)
19. From the diagram,


 is the angle of incidence
at the second face. Hence for
TIR,   1 / Sin     10 / 8    1.25
Given,  = 1.2 + 0.8 × 10 – 14 / 2.
Putting,  = 400 nm = 400 × 10 – 9 = 4 × 10 – 7 .
 1 = 1.2 + 0.8 × 10 – 14 / 16 × 10 – 14
 1 = 1.2 + 0.8/16 = 1.2 + 0.05 = 1.25.
Now for  = 500 nm = 500 × 10 – 9 = 5× 10 – 7
 2 = 1.2 + 0.8 × 10 – 14 / 25 × 10 – 14
 2 = 1.2 + 0.8/25 = 1.2 + 0.03 = 1.23
Hence only 400nm will be totally internally
reflected ( ¡ mß §çýwv 400nm §ýL â§ýʽà Ñã qå½àê ¡ àÂmáʧý
qÊàwâmêm Ñàçªàã)
àã
(A)
20. jåÞâ§ý âZà³t kv tçÞ »äþràçuà ªàuà Ñè, ¡ mß
wg = Sin [ ( A + m ) / 2 ] / Sin A/2
 wg = Sin 450 / Sin 300 = 2, hence critical
angle, Sin C = 1 / wg = Sin C = 1 / 2
 C = 450.
(B)
21. Dispersive Power âwÕ àçq½à Õ àtmà §ýL qáÊsàxàyç
 = v - r / y – 1 = ( 1.64 – 1.52 ) / 1.6 – 1
  = 0.12 / 0.6 = 1/5 = 0.2.
(C)
22. From A +  = i + e, 60 + 42 = 50 + e
 e = 520. Hence the angle made by the exit ray
with the second face ( ¡ mß âÂàªàêmî â§ýʽà õàÊà ÀåyÊç Äýv§ý yç
rÂààuà ªàuà §ýàç½à ) = 90 – 52 = 380.
(C)
23.  = Sin [ ( A + m ) / 2 ] / Sin A/2
  = Sin 450 / Sin 300 = 2 = 1.414. Hence by
definition, 12 = v1 / v2  vg = va / 
 vg = 3 × 10 10 / 2  vg = 32 × 10 10 / 2
 vg = 1.5 × 1.414 × 10 10 = 2.12 × 10 10 cm/s.
(A)
24. A +  = i + e, and as the ray exit  lary to
the second face,  e = 0, and for thin prism
 = ( - 1 ) A  i + 0 = A + ( - 1 ) A
 i = A.
A.
(D)
25. For achromatic combination ( ¡ w½àê§ý yÞuàçÞªà Ñçmä)
1 / 2 = f1/f2  1 / 2 = 2 / 3
(B)
26. For this, Sin i =  Sin A = 2 × Sin 300
 Sin i = 1 / 2  i = 450.
(C)
27. Here,  = Sin i / Sin r = Sin 2A / Sin A
  = 2 Sin A Cos A / Sin A = 2 CosA.
(B)
28. by  = Sin [ ( A + m ) / 2 ] / Sin A/2
  = Sin [ A + 180 – 2A /2 ] / Sin A/2
 = Sin ( 90 – A/2 ) / Sin A/2 = Cot A/2.
(D)
29. There are two TIRs, hence minm value of  is
given by,  = 1 / Sin 450   = 2.
(B)
30. When A = m,  = 2 Cos A/2
 Cos A/2 = 1.5 / 2 = 0.75
 A/2 = Cos -10.75 = 410  A = 820.
(C)
31. For minm deviation, A = 2r = 600  r = 30
(A)
A
I
32.

y
O

D
From the diagram, O is object and I is its image,
as prism is thin,  angle = arc / radius
  = y / D  y = D, but for a thin prism
=(-1)A
A, hence the distance of the object
from the image y = (  - 1 ) AD.
(D)
33. As from qun, 28, if  = Cot A/2, then the
angle of minm deviation m = 180 – 2A.
(D)
34. Grazing angle is given by ( qæ™þ ÐqÎàJ §ýàç½à)
Sin i = SinA ( 2 – 1 ) – CosA, given
A = 600 and  = 1.5  Sin i = 0.4682
 i = Sin -1 ( 0.4682 ) = 280.
(B)
35. For a prism at any general position ( â§ýyã âZà³t
§ýL â§ýyã sã yàtàÂu ¡ wÐnà tçÞ ) A +  = i + e,
 60 +  = 55 + 46   = 410. Hence angle of
minm deviation must be less than 410,
  < 410.
(A)