Question

A circular disc of radius \(R\) (in cm) has a uniform absorption coefficient of \(1\ \text{cm}^{-1}\). Consider a single ray passing through the disc in the plane of the disc. The shortest distance from the center of the disc to the ray is \(t\) (in cm). If \(I_i\) is the intensity of the incident ray and \(I_o\) is the intensity of the transmitted ray, then \(\log\!\left(\dfrac{I_i}{I_o}\right)\) is given by ____________________.

• A. \(2\sqrt{R^{2}-t^{2}}\)
• B. \(2R\)
• C. \(1\)
• D. \(2\sqrt{R-t}\)

Hint:

- For a uniform absorber, \(I=I_0 e^{-\mu L}\). - The path length through a circle for a ray at offset \(t\) is the chord length \(2\sqrt{R^2-t^2}\).

Answer 1

The Correct Option is A

Step 1: Beer–Lambert law: \(I_o = I_i e^{-\mu L}\), where \(\mu\) is the absorption coefficient and \(L\) is the path length through the absorber. Hence \[ \log\!\left(\frac{I_i}{I_o}\right)=\mu L . \] Step 2: Here \(\mu = 1\ \text{cm}^{-1}\), so \(\log(I_i/I_o) = L\). The ray passes through the disc along a chord whose distance from the center is \(t\). The chord length is \[ L = 2\sqrt{R^2 - t^2}. \] Step 3: Therefore, \[ \log\!\left(\frac{I_i}{I_o}\right)=2\sqrt{R^2 - t^2}. \]