Home
Class 11
PHYSICS
To find the distance d over which a sign...

To find the distance `d` over which a signal can be seen clearly in foggy conditions , a railways engineer uses dimensionsal analysis and assumes that the distance depends on the mass density `rho` of the fog , intensity `( power// area) S` of the light from the signal and its frequency to `S^(1//n)`. The values of `n` is

Text Solution

Verified by Experts

The correct Answer is:
3
`L=(I)^(n)(d)^(y)(f)^(z)`
`L=(M^(1)L^(0)T^(-3))^(x)(M^(1)L^(-3))^(y)(T^(-1))^(2)`
`L=M^(x+y)L^(-3y)T^(-3xx-2)`
`-3y=1 " " x+y=0`
`y=-1/3 " " x-1/3=0`
`x=1/3`
`L=(I)^(1//3)(d)^(-1//3)(f)^(2)`
`n=3`
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS

    ALLEN|Exercise Exercise-02|77 Videos

  • MISCELLANEOUS

    ALLEN|Exercise Exercise-03|1 Videos

  • MISCELLANEOUS

    ALLEN|Exercise EXERCISE-5(A)|15 Videos

  • KINEMATICS (MOTION ALONG A STRAIGHT LINE AND MOTION IN A PLANE)

    ALLEN|Exercise BEGINNER S BOX-7|8 Videos

  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN|Exercise EXERCISE-IV|8 Videos

Similar Questions

Explore conceptually related problems

To find the distance d over which a signal can be seen clearly in foggy conditions, a railways-engineer uses dimensions and assumes that the distance depends on the mass density rho of the fog, intensity (power/area) S of the light from the signal and its frequency f . the engineer finds that d is proportional to S^(1//n) . the value of n is

The fixed non-conducting cylinder shown in figure has a noconducting heavy piston of mass M that can slide without friction. The area of piston is S and the cylinder is filled with an ideal gas (lambda=1.5) with an initial volume V and an initial pressure P . Assume that the outside pressure on the piston is zero (vaccum). (Neglect acceleration due to gravity). For the temperature of the gas to drop to one half of its original value. the piston will have to move by a distance:

Passage XIV) A uniform cylindrical block of mass 2M and cross-sectional area A remains partially submerged in a non viscous liquid of density rho , density of the material of the cylinder is 3rho . The cylinder is connected to lower end of the tank by means of a light spring of spring constant K. The other end of the cylinder is connected to anotehr block of mass M by means of a light inextensible sting as shown in the figure. The pulleys shown are massless and frictionless and assume that the cross-section of the cylinder is very small in comparison to that of the tank. Under equilibrium conditions, half of the cylinder is submerged. [given that cylinder always remains partially immersed) If the cylinder is pushed down from equilibrium by a distance which is half the distance as calculated in the above question, determine time period of subsequent motion.

In a YDSE using monochromatic visible light, the distance between the plate of slits and the screen is 1.7 m. At point P on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved 50 cm closer to the plane of slits. Point P now lies between third and fouth minima above the central maxima and the intensity at P in one-fourth of the maxima intensity on the screen. Find the value of n.

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3 . For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. A person can see objects clearly from distance 10cm to oo . Then, we can say that the person is

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3 . For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. A near sighted man can clearly see ojects only upto a distance of 100cm and not beyond this. The number of the spectacle lenses necessary for the remedy of this defect will be

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3 . For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. A far sighted man connot see objects clearly unless they are at least 100cm from his eyes. The number of the spectacle lenses that will make his range of clear vision equal to an average grown up person will be

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3 . For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. Minimum focal length of eye lens of a normal peson is

The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained, the curvature of lens increases. That means radius of curvature decreases and focal length decreases. For a clear vision, the image must be on the retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye lens. It is about 2.5cm for a grown up person. A perosn can theoretically have clear vision of an object situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown up person, minimum distance of the object should be around 25cm. A person suffering from eye defects uses spectacles (eye glass). The function of lens of spectacles is to form the image of the objects within the range in which the person can see clearly. The image o the spectacle lens becomes object for the eye lens and whose image is formed on the retina. The number of spectacle lens used for th eremedy of eye defect is decided by the power fo the lens required and the number of spectacle lens is equal to the numerical value of the power of lens with sign. For example, if power of the lens required is +3D (converging lens of focal length 100//3cm ), then number of lens will be +3 . For all the calculations required, you can use the lens formula and lensmaker's formula. Assume that the eye lens is equiconvex lens. Neglect the distance between the eye lens and the spectacle lens. Q. Maximum focal length of a eye lens of a normal person is

ALLEN-MISCELLANEOUS-Exersice -05(B)
  1. Planck's constant has dimensions …………….

    Text Solution

    |

  2. In the formula x=3yz^(2), x and y have dimensions of capacitance and m...

    Text Solution

    |

  3. The dimensions of electrical conductivity are ……..

    Text Solution

    |

  4. The equation of state of a real gas is given by (P+a/V^(2)) (V-b)=RT ...

    Text Solution

    |

  5. The dimensions of 1/2 epsilon(0)E^(2) (epsilon(0)= permittivity of fre...

    Text Solution

    |

  6. A quantity X is given by epsi(0)L(DeltaV)/(DeltaT) where epsi(0) is th...

    Text Solution

    |

  7. The relation p=(alpha)/(beta)e^((-alphaZ)/(ktheta) , where p is pressu...

    Text Solution

    |

  8. Which of the following sets have different dimensions ?

    Text Solution

    |

  9. L,C, and R represent the physical quantities inductance, capacitance a...

    Text Solution

    |

  10. The dimensions of the quantities in one ( or more) of the following pa...

    Text Solution

    |

  11. The pairs of physical quantities that have the same dimensions is (are...

    Text Solution

    |

  12. Let varepsilon(0) denote the permittivity of the vacuum and mu(0) that...

    Text Solution

    |

  13. The SI unit of inductance, the henry, can be written as

    Text Solution

    |

  14. Column-1 gives three physical quantities. Select the appropriate units...

    Text Solution

    |

  15. Match the physical quantities given in column I with dimensions expres...

    Text Solution

    |

  16. Ratio of amplitude for two wave is 15:9.Find the ratio of intensity?

    Text Solution

    |

  17. Match List I with List II and select the correct answer using the code...

    Text Solution

    |

  18. During Searle's experiment zero of the Vernieer scale lies between 3.2...

    Text Solution

    |

  19. To find the distance d over which a signal can be seen clearly in fogg...

    Text Solution

    |