Problem 1 - Master Resource Book in JEE Main Physics

Question

Two point sources \(A\) and \(B\) of luminous intensities 1 od and 16 cd respectively are placed \(100 \mathrm{~cm}\) apart. A \mathrm{\\{} g r e a s e ~ s p o t ~ s e r e e n ~ i s ~ p l a c e d ~ b e t w e e n ~ t h e ~ t w o ~ s o u r c e s . ~ For the grease spot to become indistinguishable from both the sides, it should be placed at (a) \(80 \mathrm{~cm}\) from 16 cd lamp and \(20 \mathrm{~cm}\) from \(1 \mathrm{~cd}\) (b) \(20 \mathrm{~cm}\) from the \(16 \mathrm{~cd}\) and \(80 \mathrm{~cm}\) from \(1 \mathrm{~cd}\) (c) \(\frac{400}{3} \mathrm{~cm}\) from \(16 \mathrm{~cd}\) and \(\frac{100}{3} \mathrm{~cm}\) from \(1 \mathrm{~cd}\) (d) \(\frac{100}{3} \mathrm{~cm}\) from \(16 \mathrm{~cd}\) and \(\frac{400}{3} \mathrm{~cm}\) from \(1 \mathrm{~cd}\)

Step-by-Step Solution

Answer

The grease spot should be \(20 \mathrm{~cm}\) from the \(16 \mathrm{~cd}\) lamp and \(80 \mathrm{~cm}\) from the \(1 \mathrm{~cd}\) lamp (option b).

Step 1: Understand the Problem

For the grease spot on a screen to become indistinguishable from both light sources, the illuminance on the screen from both sources must be equal. We have two light sources, \( A \) with an intensity of \( 1 \text{ cd} \) and \( B \) with an intensity of \( 16 \text{ cd} \), placed \( 100 \text{ cm} \) apart.

Step 2: Use the Inverse Square Law

The illuminance \( E \) on a screen at a distance \( r \) from a point source of intensity \( I \) is given by \( E = \frac{I}{r^2} \). We need \( E_A = E_B \) for the grease spot to be indistinguishable.

Step 3: Set Up the Equation for Equal Illuminance

Let the distance of the grease spot from \( A \) be \( x \). Thus, the distance from \( B \) will be \( 100 - x \). The condition for equal illuminance is: \[ \frac{1}{x^2} = \frac{16}{(100-x)^2} \]

Step 4: Solve the Equation for \( x \)

Cross multiply to solve the equation: \[ (100-x)^2 = 16x^2 \] Expanding and simplifying gives: \[ 10000 - 200x + x^2 = 16x^2 \] \[ 15x^2 + 200x - 10000 = 0 \] Divide the equation by 5: \[ 3x^2 + 40x - 2000 = 0 \]

Step 5: Solve the Quadratic Equation

Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( 3x^2 + 40x - 2000 = 0 \). Here, \( a = 3 \), \( b = 40 \), \( c = -2000 \). The solution(s): \[ x = \frac{-40 \pm \sqrt{1600 + 24000}}{6} \] \[ x = \frac{-40 \pm 160}{6} \] This gives two possible solutions: \( x = 20 \) cm and \( x = -100/3 \) cm, discard the negative solution.

Step 6: Verify Acting Solution Against Choices

From \( x = 20 \) cm, the distance from \( B \) is \( 80 \) cm. This matches option (b). Thus, the grease spot is \( 20 \mathrm{~cm} \) from the \( 16 \mathrm{~cd} \) source and \( 80 \mathrm{~cm} \) from the \( 1 \mathrm{~cd} \) source.

Key Concepts

Luminous Intensity

Luminous Intensity is a measure of the amount of visible light emitted in a particular direction by a light source. It is measured in candelas (cd). Consider a bulb that emits light evenly in all directions. Each part of the light emitted is perceived as luminous intensity. The concept is crucial in understanding how much light is directed towards a specific point or area and helps in calculating the brightness seen from different perspectives. In our scenario, the light sources are characterized by their respective luminous intensities, which are 1 cd and 16 cd. This difference shows that the second light source emits significantly more light in the direction of the grease spot than the first. This becomes important when striving for equal illuminance from both sources.

Point Sources

Point Sources are idealized sources of light that emit light in all directions from a single point. In physics and practical calculations, treating a light as a point source simplifies modeling how light spreads over distances. Each source in this exercise, emitting light, is considered to be a point source. This simplification means we can apply straightforward mathematical principles such as the Inverse Square Law. The Inverse Square Law explains how the light intensity diminishes as it moves away from the source. In the given problem, each of the two light sources is at a specific point, providing a clear path to calculate the influence of each light source on the grease spot.

Quadratic Equation

Quadratic Equation is a fundamental mathematical concept where the highest power of the variable is squared (second degree). It takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. Identifying and solving these equations is critical in physics because they help find unknown distances, such as in this exercise. In our case, the method to equal the illuminance from two different light sources requires solving a quadratic equation. By setting up the equation \( 3x^2 + 40x - 2000 = 0 \), we find out where, along the 100 cm path, the illuminance from both sources will be equal. Solving this equation allows us to pinpoint the optimal position for the grease spot.

Illuminance

Illuminance is the measure of how much luminous flux is spread over a given area. It is measured in lux (lx) and is calculated by dividing the luminous intensity by the square of the distance from the source, according to the relation \( E = \frac{I}{r^2} \). This concept tells us how the brightness of a light source decreases as we move away from it. For the problem, both light sources must produce the same illuminance on the grease spot. Thus, given the differences in their intensities, the position of the grease spot is adjusted such that the brightness from both sources appears equal at this point. This makes the term 'indistinguishable' relevant in determining the precise location for the grease spot to achieve equal illuminance from both points of light.