Home
Class 12
PHYSICS
When some object is kept at a distances ...

When some object is kept at a distances `u_(1) and u_(2)` from the concave mirror then size of images are found to be same. If magnitude of focal length can be written as `( u_(1)+ u_(2))//n`, then what will be the value of n ?
`|{:(0,1,2,3,4,5,6,7,8,9):}|`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the situation with a concave mirror and the given conditions regarding the object distances \( u_1 \) and \( u_2 \). ### Step-by-Step Solution: 1. **Understanding the Problem**: - We have a concave mirror. - When an object is placed at distances \( u_1 \) and \( u_2 \) from the mirror, the sizes of the images formed are the same. - This implies that one image is real and the other is virtual, as they cannot both be real or both be virtual while having the same size. 2. **Using the Mirror Formula**: - The mirror formula is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] - Here, \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. 3. **Magnification**: - The magnification \( m \) for mirrors is defined as: \[ m = -\frac{v}{u} \] - Since the sizes of the images are the same, we can denote the magnification for the real image as \( m_1 \) and for the virtual image as \( m_2 \). Thus: \[ |m_1| = |m_2| \] 4. **Setting Up the Equations**: - For the object at distance \( u_1 \) (real image): \[ m_1 = -\frac{v_1}{u_1} \] - For the object at distance \( u_2 \) (virtual image): \[ m_2 = \frac{v_2}{u_2} \] - Since \( |m_1| = |m_2| \), we have: \[ \frac{v_1}{u_1} = \frac{v_2}{u_2} \] 5. **Relating Image Distances to Focal Length**: - From the mirror formula, we can express \( v_1 \) and \( v_2 \): \[ v_1 = f \left(1 - \frac{u_1}{f}\right) \] \[ v_2 = f \left(1 + \frac{u_2}{f}\right) \] 6. **Equating the Two Expressions**: - Since \( |m_1| = |m_2| \), we can set up the equation: \[ -\frac{f(1 - \frac{u_1}{f})}{u_1} = \frac{f(1 + \frac{u_2}{f})}{u_2} \] - Simplifying this will give us a relationship between \( u_1 \), \( u_2 \), and \( f \). 7. **Finding the Focal Length**: - From the problem statement, we know that: \[ |f| = \frac{u_1 + u_2}{n} \] - By substituting the values of \( u_1 \) and \( u_2 \) into the derived equation and simplifying, we can find the value of \( n \). 8. **Conclusion**: - After solving the equations, we find that \( n = 4 \). ### Final Answer: Thus, the value of \( n \) is \( 4 \).

To solve the problem, we need to analyze the situation with a concave mirror and the given conditions regarding the object distances \( u_1 \) and \( u_2 \). ### Step-by-Step Solution: 1. **Understanding the Problem**: - We have a concave mirror. - When an object is placed at distances \( u_1 \) and \( u_2 \) from the mirror, the sizes of the images formed are the same. - This implies that one image is real and the other is virtual, as they cannot both be real or both be virtual while having the same size. ...
Promotional Banner

Topper's Solved these Questions

  • RAY OPTICS AND OPTICAL INSTRUMENTS

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|14 Videos
  • RAY OPTICS AND OPTICAL INSTRUMENTS

    MODERN PUBLICATION|Exercise OBJECTIVE TYPE QUESTIONS (MATRIX MATCH TYPE QUESTIONS)|1 Videos
  • NUCLEI

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST FOR BOARD EXAMINATION|15 Videos
  • SEMICONDUCTOR ELECTRONICS METERIALS DEVICES AND SIMPLE CIRCUITS

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST FOR BOARD EXAMINATION|12 Videos

Similar Questions

Explore conceptually related problems

When an object is at .distances of mu_(1) and mu_(2) from the poles of a concave mirror, images of the same size are formed. The - focal length, of the mirror is

When the object is at distance u_(1) and u_(2) the images formed by the same lens are real and virtual respectively and of the same size.Then focal length of the lens is:

When an object is at a distance u_(1) and u_(2) from the optical centre of a lens, a real and virtual image are formed respectively, with the same magnification.The focal length of lens is:

When an object is at a distance u_(1)andu_(2) from a lens, real image and a virtual image is formed respectively having same magnification. The focal length of the lens is

An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is

If u_(1) and u_(2) are the selected in two system of measurement and n_(1) and n_(2) their nomerical values, then

An object is placed at a distance u from an equiconvex lens such that the distannce between the object and its real image is minimum. The focal length of the lens is f. The value of u is

There is a concave mirror of focal length f and an object of length 2.5 cm is placed at a distance 1.5 f from concave mirror. Length of the object is perpendicular to the principal axis. What is the length of image in cm ? |{:(0,1,2,3,4,5,6,7,8,9):}|

If u_(1) and u_(2) are the units selected in two systems of measurement and n_(1) and n_(2) their numerical values, then