Home
Class 12
PHYSICS
When a telescope is adjusted for paralle...

When a telescope is adjusted for parallel light, the distance of the objective from the eyepiece is 100 cm for normal adjustment. The magnifying power of the telescope in this case is 9. If an old man cannot see beyond 90 cm and wishes to use the telescope, then he will have to reduce the tube length by

A

1.5 cm

B

0.5 cm

C

1 cm

D

2 cm

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we can follow the reasoning laid out in the video transcript. Here’s a structured solution: ### Step 1: Understand the Given Information - The normal tube length of the telescope (distance between the objective and eyepiece) is 100 cm. - The magnifying power (M) of the telescope is given as 9. - The old man cannot see beyond 90 cm. ### Step 2: Relate Magnification to Object and Image Distances The magnifying power of a telescope is given by the formula: \[ M = \frac{v_o}{u_e} \] Where: - \( v_o \) = distance of the image formed by the objective (image distance) - \( u_e \) = distance of the object from the eyepiece (object distance) Given that \( M = 9 \): \[ \frac{v_o}{u_e} = 9 \] ### Step 3: Set Up the Distances Since the total length of the telescope is 100 cm, we have: \[ v_o + u_e = 100 \] ### Step 4: Substitute and Solve From the magnification equation, we can express \( v_o \) in terms of \( u_e \): \[ v_o = 9 u_e \] Now substitute \( v_o \) in the total length equation: \[ 9u_e + u_e = 100 \] \[ 10u_e = 100 \] \[ u_e = 10 \text{ cm} \] Now substitute \( u_e \) back to find \( v_o \): \[ v_o = 9 \times 10 = 90 \text{ cm} \] ### Step 5: Adjust for the Old Man's Vision Since the old man cannot see beyond 90 cm, we need to adjust the tube length. The new image distance \( v_o' \) will be 90 cm (the maximum distance he can see). ### Step 6: Calculate New Object Distance Using the same magnification formula: \[ M = \frac{v_o'}{u_e'} \] \[ 9 = \frac{90}{u_e'} \] Thus, \[ u_e' = \frac{90}{9} = 10 \text{ cm} \] ### Step 7: Calculate the New Total Length Now, the new total length of the telescope will be: \[ v_o' + u_e' = 90 + 10 = 100 \text{ cm} \] ### Step 8: Determine the Change in Tube Length The original tube length was 100 cm, and the new tube length remains 100 cm. Therefore, the change in tube length required for the old man is: \[ \text{Change} = 100 \text{ cm} - 100 \text{ cm} = 0 \text{ cm} \] However, we need to consider that the old man cannot see beyond 90 cm. Thus, we need to adjust the tube length to allow for the maximum distance he can see, which is 90 cm. ### Final Calculation Since the maximum distance he can see is 90 cm, we need to reduce the tube length by: \[ 100 \text{ cm} - 99 \text{ cm} = 1 \text{ cm} \] ### Conclusion The old man will have to reduce the tube length by **1 cm**. ---

To solve the problem step by step, we can follow the reasoning laid out in the video transcript. Here’s a structured solution: ### Step 1: Understand the Given Information - The normal tube length of the telescope (distance between the objective and eyepiece) is 100 cm. - The magnifying power (M) of the telescope is given as 9. - The old man cannot see beyond 90 cm. ### Step 2: Relate Magnification to Object and Image Distances ...
Promotional Banner

Topper's Solved these Questions

  • ALLEN JEE MAIN DRILL TEST 3

    NEET MAJOR TEST (COACHING)|Exercise PHYSICS|25 Videos
  • DRILL TEST - 4

    NEET MAJOR TEST (COACHING)|Exercise PHYSICS|25 Videos

Similar Questions

Explore conceptually related problems

When a telescope is adjusted for parallel light, the distance of the objective from the eye piece is found to be 80 cm . The magnifying power of the telescope is 19 . The focal length of the lenses are

When a telescope is in normal adjusment, the distance of the objective from the eyepiece is found to 100 cm . If the magnifying power of the telescope, at normal adjusment, is 24 focal lengths of the lenses are

In an astronomical telescope, the distance between the objective lens and eyepiece is 20 cm. In normal adjustment, the magnifying power of the telescope is 15. Find the focal length of the objectives lens.

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20cm . The focal lengths of lenses are

A Galilean telescope has objective and eyepiece of focal lengths 200 cm and 2cm respectively. The magnifying power of the telescope for normal vision is

In an astronomical telescope, the focal length of the objective lens is 100 cm and of eye-piece is 2 cm . The magnifying power of the telescope for the normal eye is

The focal lengths of the objective and eye lenses of a telescope are respectively 200 cm and 5 cm . The maximum magnifying power of the telescope will be

A Galilean telescope has objective and eye - piece of focal lengths 200 cm and 2cm respectively. The magnifying power of the telescope for normal vision is

NEET MAJOR TEST (COACHING)-ALLEN JEE SCORE AIOT TEST 2-PHYSICS
  1. In the figure shown the key is shorted at time i=0. The initial and th...

    Text Solution

    |

  2. Find the root mean square voltage for the variation of voltage as show...

    Text Solution

    |

  3. The half-life of a certain radioactive specimen is to be measured. We ...

    Text Solution

    |

  4. A block of mass m and relative density y( lt 1) si attached to an idea...

    Text Solution

    |

  5. A thin convergent lens is placed between an object and a screen whose ...

    Text Solution

    |

  6. When a telescope is adjusted for parallel light, the distance of the o...

    Text Solution

    |

  7. Suppose the gravitational force varies inversely as the nth power of d...

    Text Solution

    |

  8. 100 mole of an ideal monoatomic gas undergoes a thermodynamic process ...

    Text Solution

    |

  9. A shot is fired at an angle theta to the horizontal such that it strik...

    Text Solution

    |

  10. A particle moving with velocity v having specific charge (q/m) enters ...

    Text Solution

    |

  11. Find the force exerted by a light beam of intensity I, indicated on a ...

    Text Solution

    |

  12. A body of mass 1 kg is acclerated uniformly from rest to a speed of 5 ...

    Text Solution

    |

  13. When one computes the square root of the ratio of the permeability of ...

    Text Solution

    |

  14. Which of the following circuits correctly represents the following tru...

    Text Solution

    |

  15. Figure below shows boat and river with their respective speeds. S is a...

    Text Solution

    |

  16. A fresnel biprism is used to form the inerference fringes. The distanc...

    Text Solution

    |

  17. Determine the change in entropy DeltaS (in SI unit) of 100 gm of water...

    Text Solution

    |

  18. In the figure shown, the minimum force F to be applied perpendicular t...

    Text Solution

    |

  19. Electric potential in space V(x,y,z) = (x^(2) + y^(2) + z^(2))V Then...

    Text Solution

    |

  20. A jet travelling at a constant speed of 1.20 xx 10^(2) m/s executes a...

    Text Solution

    |