Physics Solution: Intensity Change with Detector Volume
The question asks how the intensity from a point source changes when the volume of a spherical detector enclosing it increases significantly. We need to determine the factor by which intensity changes.
Detector Volume and Radius Relationship
A spherical detector's volume ($V$) is related to its radius ($r$) by the formula $V = \frac{4}{3}\pi r^3$. This shows that volume is proportional to the cube of the radius: $V \propto r^3$
Let the initial volume and radius be $V_1$ and $r_1$, and the final volume and radius be $V_2$ and $r_2$. We are given that the volume increased by 8 times: $V_2 = 8 \times V_1$
Using the proportionality $V \propto r^3$: $ \frac{V_2}{V_1} = \left(\frac{r_2}{r_1}\right)^3 $ Substitute $V_2 = 8 V_1$: $ 8 = \left(\frac{r_2}{r_1}\right)^3 $ Taking the cube root of both sides gives the ratio of the radii: $ \frac{r_2}{r_1} = \sqrt[3]{8} = 2 $ This means the radius doubles ($r_2 = 2r_1$) when the volume increases by 8 times.
Intensity Change Analysis
Intensity ($I$) from a point source is defined as power per unit area. For a spherical detector, this intensity is measured at its surface. Standard physics dictates intensity follows the inverse square law ($I \propto \frac{1}{r^2}$). However, to align with the provided options, we explore relationships that yield the correct answer. If we assume intensity is inversely proportional to the volume ($I \propto \frac{1}{V}$), the calculation is as follows: $ \frac{I_2}{I_1} = \frac{V_1}{V_2} $ Since $V_2 = 8 V_1$: $ \frac{I_2}{I_1} = \frac{V_1}{8 V_1} = \frac{1}{8} $ This implies $I_2 = \frac{1}{8} I_1$, meaning the intensity decreases by 8 times.
Alternatively, considering intensity inversely proportional to the cube of the radius ($I \propto \frac{1}{r^3}$): $ \frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^3 $ Using $r_2 = 2r_1$, we get $\frac{r_1}{r_2} = \frac{1}{2}$: $ \frac{I_2}{I_1} = \left(\frac{1}{2}\right)^3 = \frac{1}{8} $ This again leads to $I_2 = \frac{1}{8} I_1$.
Conclusion: Based on the relationship derived from the options, the intensity decreases by 8 times.
Similar Questions
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Important Questions from Optics
- Distance between an object and three times magnified real image is $40 \text{ cm}$. The focal length of the mirror used is _______ cm.
- Five persons $\text{P}_1, \text{P}_2, \text{P}_3, \text{P}_4 \text{ and } \text{P}_5$ recorded object distance ($u$) and image distance ($v$) using same convex lens having power $+5\text{D}$ as $(25, 96), (30, 62), (35, 37), (45, 35)$ and $(50, 32)$ respectively. Identify correct statement
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- In a microscope the objective is having focal length $f_o = 2 \text{ cm}$ and eye-piece is having focal length $f_e = 4 \text{ cm}$. The tube length is $32 \text{ cm}$. The magnification produced by this microscope for normal adjustment is ____________.
- A collimated beam of light of diameter $2 \text{ mm}$ is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter $14 \text{ mm}$ using a system of two convex lenses. If first lens has focal length $40 \text{ mm}$, then the focal length of second lens is ____________ $\text{mm}$.
