A ray of light is incident at `60^(@)` on one face of a prism of angle `30^(@)` and the emergent ray makes `30^(@)` with the incident ray. The refract

Monjur Alahi

A ray of light is incident at `60^(@)` on one face of a prism of angle `30^(@)` and the emergent ray makes `30^(@)` with the incident ray. The refractive index of the prism is
A. 1.732
B. 1.414
C. 1.5
D. 1.33


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A
(a) As, A+ `delta`=i+e
`rArr 30^(@) + 30^(@) = 60^(@)` + e
Angle of emergence , e=`60^(@)-60^(@)`=0
`r_(1)=A=30^(@)`
`therefore` Refractive index, `mu=(sin i)/(sin r)=sin 60^(@)/(sin 30^(@))=sqrt3=1.732`

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
A ray of light is incident at an angle of `60^(@)` on one face of a prism of angle `30^(@)`. The ray emerging out of the prism makes an angle of `30^(

Correct Answer - A (a) Deviation of ray in prism is given by, `delta=i+e-A` So, `e=delta+A-i=30^(@)-60^(@)=0^(@)` So, emergent ray will be perpendicular to face, or emergent ray will make an angle...

2 Answers 1 views
Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism The angle of prism is `(cos 41^(@) = 0.75)`

Correct Answer - D (d) Refractive index of prism, µ`=(sin""((A+delta_(m))/2))/(sin(A/2))` substituting `A=delta_(m)and µ=1.5,` `1.5=(sinA)/(sin((A/2)))` `(1.5)/2=cos""A/2` `7.5=cos""A/2` We get, A=`82^(@)`.

2 Answers 1 views
When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism `=1.5`

Correct Answer - D (d) Here, `i_(1)=0=r_(1)` Therefore,`r_(2)=A=60^(@)` `theta_(C)=sin^(-1)(1/1.5)=41.8^(@)` `rArr r_(2)gttheta_(C)`

2 Answers 1 views
A glass prism has a refractive angle of `90^(@)` and a refractive index of 1.5. A ray is incident at an angle of `30^(@)`. The ray emerges from an adj

Correct Answer - D (d) Critical angle, `theta_(C)=sin^(-1)(1/1.5)=41.8^(@)` Incident angle, `i_(1)=30^(@)` `sinr_(1)=(sini_(1))/(mu)=(sin(30^(@)))/(1.5)` `r_(1)=19.5^(@)` or Angle of emergence, `r_(2)=A-r_(1)=90-19.5=70.5^(@)` Since, `r_(2)gttheta_(C)` `therefore` The ray will not emerge from the prism.

2 Answers 1 views
One face of a prism with a refrective angle of `30^@` is coated with silver. A ray of light incident on another face at an angle of `45^@` is refracte

Correct Answer - A (a) Given, `r_(2)=0,r_(1)=A=30^(@),and i_(1)=45^(@)` `therefore` Refractive index, `mu=sini_(1)/sinr_(1)=sin45^(@)/sin30^(@)=(1//sqrt2)/(1//2)=sqrt2`

2 Answers 1 views
When light is incident on a glass slab, the incident ray, refracted ray and the emergent ray are in three media A, B and C.

Correct answer is (d) n1 = n3 < n2  Medium A and C in air, therefore, n1 = n3, but glass slab is of higher refractive index. ∴ n1 = n3 < n2

2 Answers 1 views
Assertion angle of deviation depends on the angle of prism. Reason For thin prism =`(mu-1)A` Where =angle of deviation `mu=refractive index, A=angle o

Correct Answer - A The relation between angle of dieviation `delta` for a thin prism an angle of prism and refractive index `(mu)` of material of prism is given buy `delta...

2 Answers 1 views
A beam of monochromatic light is incident at `i= 50^(@)` on one face of an equilateral prism, the angle of emergence is `40^(@)`, then the angle of mi

Correct Answer - B `I = 50^(@)` , equilateral prism `A = 60^(@) , e= 40^(@)` Total angle of deviation `delta =i+e - A = 50 + 40 - 60, delta=...

2 Answers 1 views
Light of wavelenght `4000 A` is incident at small angle on a prim of apex angle `4^(@)` . The prism has `n_(v)=1.5` and `n_(r)=1.48` . The angle of di

Correct Answer - C Dispersion will not occur for a monochromatic light.

2 Answers 1 views
In an isosceles prism of prism angle `45^(@)`, it is found that when the angle of incidence is same as the prism angle, the emergent ray grazes the em

Correct Answer - D `i=(pi)/(2),e=(pi)/(4),A=(pi)/(4)` `(sini)/(sin r_(1))=(sine)/(sinr_(2))=mu` `rArr sinr_(1)=(1)/(mu) and sinr_(2)=(1)/(sqrt2mu)` Since `r_(1)+r_(2)=A=(pi)/(4) rArr r_(1)=(pi)/(4)-r_(2)` `rArr sinr_(1) =(1)/(sqrt2)cosr_(2)- (1)/(sqrt2)sinr_(2)` `=sqrt2 sinr_(1)+sinr_(2)=cosr_(2)` `=(sqrt2)/(mu)+(1)/(sqrt2mu)=sqrt(1-(1)/(2mu^(2)))= (1)/(mu^(2))(2+(1)/(2)+2)=1-(1)/(2mu^(2))` `= (1)/(mu^(2))((9)/(2)+(1)/(2))=1` `mu^(2)=5rArr mu=sqrt5 `.

2 Answers 1 views
X