1. The refractive index of the material of a prism is √2 and its refracting angle is 30°. One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light enters the prism from the mirror surface if its angle of incidence of the prism is
   

1. 1. 30°

   
   
   

   2. 45°

   
   
   

   3. 60°

   
   
   

   4. 0°

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∠r = 30° (using law of triangle)
   ​​
   

   ⇒ µ =	sin i
   	sin r

   ​​ ​​
   ⇒ √2 × sin 30° = sin i
   

   ⇒ sin i =	1	⇒ i = 45°
   	√2

   Correct Option: B

   ∠r = 30° (using law of triangle)
   ​​
   

   ⇒ µ =	sin i
   	sin r

   ​​ ​​
   ⇒ √2 × sin 30° = sin i
   

   ⇒ sin i =	1	⇒ i = 45°
   	√2

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2. A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance :
   

1. 1. 37.3 cm

   
   
   

   2. 46.0 cm

   
   
   

   3. 50.0 cm

   
   
   

   4. 54.0 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given: Focal length of objective, f₀ = 40cm
   ​Focal length of eye – piece f_e = 4 cm
   ​image distance, v₀ = 200 cm ​
   Using lens formula for objective lens ​
   

   	1	-	1	=	1	⇒	1	=	1	-	1
   	v0		u0		ƒ0		v0		ƒ0		u0

   
   

   ⇒	1	=	1	+	1	=	+ 5 - 1
   	v0		40		- 200		200

   
   ⇒​v₀ = 50 cm ​
   Tube length ℓ = |v₀| + f_e = 50 + 4 = 54 cm.

   Correct Option: D

   Given: Focal length of objective, f₀ = 40cm
   ​Focal length of eye – piece f_e = 4 cm
   ​image distance, v₀ = 200 cm ​
   Using lens formula for objective lens ​
   

   	1	-	1	=	1	⇒	1	=	1	-	1
   	v0		u0		ƒ0		v0		ƒ0		u0

   
   

   ⇒	1	=	1	+	1	=	+ 5 - 1
   	v0		40		- 200		200

   
   ⇒​v₀ = 50 cm ​
   Tube length ℓ = |v₀| + f_e = 50 + 4 = 54 cm.

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3. Angle of deviation (δ) by a prism (refractive index = µ and supposing the angle of prism A to be small) can be given by
   

1. 1. δ = (μ - 1) A

   
   
   

   2. δ = (μ + 1) A

   
   
   

   3. δ =	sin	A + δ
      		2
      	sin	A
      		2

   
   
   

   4. δ =	µ - 1	A
      	µ + 1

   
   
   

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1. View Hint View Answer Discuss in Forum

   When the angle of prism is small, δ = (µ – 1) A

   Correct Option: A

   When the angle of prism is small, δ = (µ – 1) A

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4. The refractive index of the material of the prism is √3 ; then the angle of minimum deviation of the prism is
   

1. 1. 30°

   
   
   

   2. 45°

   
   
   

   3. 60°

   
   
   

   4. 75°

   
   
   

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1. View Hint View Answer Discuss in Forum

   Angle of minimum deviation ​ ​​ ​
   

   μ =	sin		A + δm			⇒ √3 =	sin		A + δm
   			2						2
   	sin		A				sin		60°
   			2						2

   
   

   ⇒ sin		30° +	δm		=	√3	⇒ 30 °	δm	= 60°
   			2			2		2

   
   ⇒ δ_m = 60°.

   Correct Option: C

   Angle of minimum deviation ​ ​​ ​
   

   μ =	sin		A + δm			⇒ √3 =	sin		A + δm
   			2						2
   	sin		A				sin		60°
   			2						2

   
   

   ⇒ sin		30° +	δm		=	√3	⇒ 30 °	δm	= 60°
   			2			2		2

   
   ⇒ δ_m = 60°.

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5. In an astronomical telescope in normal adjustment a straight black line of lenght L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is l. The magnification of the telescope is :
   

1. 1. 	L	- 1
      	I

      
      

   
   
   

   2. 	L + I
      	L - I

   
   
   

   3. 	L
      	I

   
   
   

   4. 	L	+ l
      	I

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Magnification by eye piece ​m = ​ f/f + u
   

   -	I	=	fe	= -	fe	or,	I	=	fe
   	L		fe + I - (f0 + fe)I		f0		L		f0

   
   

   Magnification, M =	f0	=	L
   	fe		I

   Correct Option: C

   
   Magnification by eye piece ​m = ​ f/f + u
   

   -	I	=	fe	= -	fe	or,	I	=	fe
   	L		fe + I - (f0 + fe)I		f0		L		f0

   
   

   Magnification, M =	f0	=	L
   	fe		I

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