6. For a normal eye, the cornea of eye provides a converging power of 40D and the least converging power of the eye lens behind the cornea is 20D. Using this information, the distance between the retina and the eye lens of the eye can be estimated to be
   

1. 1. 2.5 cm

   
   
   

   2. 1.67 cm

   
   
   

   3. 1.5 cm

   
   
   

   4. 5 cm

   
   
   

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1. View Hint View Answer Discuss in Forum

   P_cornea = + 40 D ​
   P_e = + 20 D ​
   Total power of combination = 40 + 20 = 60 D ​
   

   Focal length of combination =	1	× 100 =	5	cm
   	60		3

   ​ ​
   For minimum converging state of eye lens, ​ ​
   

   u = -∞ V = ? f =	5
   	3

   
   

   From lens formula, =	1	=	1	-	1	⇒ v =	5	cm
   	f		v		u		3

   ​
   ​Distance between retina and cornea-eye lens ​
   

   =	5	= 1.67 m
   	3

   Correct Option: B

   P_cornea = + 40 D ​
   P_e = + 20 D ​
   Total power of combination = 40 + 20 = 60 D ​
   

   Focal length of combination =	1	× 100 =	5	cm
   	60		3

   ​ ​
   For minimum converging state of eye lens, ​ ​
   

   u = -∞ V = ? f =	5
   	3

   
   

   From lens formula, =	1	=	1	-	1	⇒ v =	5	cm
   	f		v		u		3

   ​
   ​Distance between retina and cornea-eye lens ​
   

   =	5	= 1.67 m
   	3

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7. If the focal length of objective lens is increased then magnefying power of :
   

1. 1. microscope will increase but that of telescope decrease.

   
   
   

   2. microscope and telescope both will increase.

   
   
   

   3. microscope and telescope both will decrease

   
   
   

   4. microscope will decrease but that of telescope increase.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Magnifying power of microscope
   

   =	LD	∝	1
   	fef0		f0

   
   ​Hence with increase f₀ magnifyig power of microscope decreases. ​
   

   Magnifying power of telescope =	f0	∝ f0
   	fe

   ​
   Hence with increase f₀ magnifying power of telescope increases.

   Correct Option: D

   Magnifying power of microscope
   

   =	LD	∝	1
   	fef0		f0

   
   ​Hence with increase f₀ magnifyig power of microscope decreases. ​
   

   Magnifying power of telescope =	f0	∝ f0
   	fe

   ​
   Hence with increase f₀ magnifying power of telescope increases.

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8. 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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9. 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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10. 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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