Suppose Ram has placed a ball in front of a concave mirror of focal length (f )at various distances (u) and he has measured the corresponding image distances (v). From the values of u and v, Ram is able to plot a graph of 1/v against 1/u. From the graph, Ram has observed that x intercept and y intercept are 0.1" " cm and 0.11" " cm, respectively. Then the focal length of the mirror is:

Correct option is C

Correct Answer: C 9.55 cm

Explanation:

The relationship between the object distance (uuu), image distance (vvv), and focal length (fff) of a concave mirror is described by the mirror formula:

​$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$​​

Rewriting it for graph plotting:

​$$\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$$​​

Here:

• $$\frac{1}{v}$$​​ is plotted along the y-axis,
• $$\frac{1}{u}$$​​ is plotted along the x-axis,
• The slope m=−1,
• The y-intercept (c) is$$\frac{1}{f}$$​.

From the problem:

• x-intercept: $$\frac{1}{u}$$​ = 0.1 cm⁻¹,
• y-intercept:$$\frac{1}{v} = 0.11 \, \text{cm}^{-1}.$$​​

The y-intercept gives $$\frac{1}{f} = 0.11 \, \text{cm}^{-1}$$​. Thus:

​$$f = \frac{1}{0.11} = 9.55 \, \text{cm}.$$​​

This value corresponds to the focal length of the mirror.

Information Booster:

•  The y-intercept $$(\frac{1}{f}​)$$​ directly determines the focal length of the mirror.
  
•  The graph's slope (−1) confirms the inverse relationship between $$\frac{1}{u}​and \frac{1}{v}​.$$​​
  
•  The x-intercept $$(\frac{1}{u} = 0.1 \, \text{cm}^{-1})$$​ implies the object distance would be u=10 cm when the image is at infinity.
  
• Concave mirrors can form both real and virtual images depending on the object's position relative to the focal point.
  
• The focal length determines the convergence strength of the mirror and is crucial in optical applications.