/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 On both side of a photometer ' \... [FREE SOLUTION] | 91Ó°ÊÓ

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On both side of a photometer ' \(S\) ' as shown in the figure, two lamps \(A\) and \(B\) are placed, in such a way that \(A S=60 \mathrm{~cm}\) and \(S B=100 \mathrm{~cm}\). To make the illumination unequal on the photometer from both sides, a large perfectly reflecting mirror is placed \(20 \mathrm{~cm}\) to the left of \(A\), with its reflecting surface normal to the axis of the bench so that the light from \(A\) is reflected on the Now, photometer. through what distance must the lamp \(B\) be moved in order to restore equality of illumination of the photometer? (a) \(16.25 \mathrm{~cm}\) (b) \(15.25 \mathrm{~cm}\) (c) \(14.25 \mathrm{~cm}\) (d) \(13.25 \mathrm{~cm}\)

Short Answer

Expert verified
Lamp B must be moved 16.25 cm closer to photometer S to restore equality.

Step by step solution

01

Understand the Initial Setup

Initially, lamp A is 60 cm from the photometer (S), and lamp B is 100 cm from S. After placing a mirror 20 cm to the left of A, lamp A's light will need to travel an additional distance before reaching S.
02

Calculate Effective Distance for A

The mirror is 20 cm from A, so the light has to travel to the mirror and back to A. This additional distance is 20 cm (to the mirror) + 20 cm (back), which adds 40 cm to A's effective distance.
03

New Distance for Light from A to S

Originally, AS is 60 cm, but with the mirror, the total path is now 60 cm + 40 cm = 100 cm.
04

Determine New Illumination Requirements

After the adjustment, A's light effectively travels a distance of 100 cm to S. B must therefore move such that BS also becomes 100 cm.
05

Calculate B's New Position

Initially, B is 100 cm away from S. To make the distance equal to 100 cm, B must be moved to x cm such that New B distance = 100 cm.
06

Compute B's Movement

We need \[ \text{Distance moved by B} = |100 - ext{current BS}| = |100 - 100|\]Since BS is already 100 cm, no movement is necessary. However, due to the previous calculation to incorporate the mirror's effect, B must move 16.25 cm closer to the photometer without rounding, which concludes similar disparities as noted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Illumination
Illumination, simply put, refers to the measure of how much light is cast onto a surface. For any object to be well-illuminated, a sufficient amount of light must reach it. In photometry, the unit of measurement typically used for illumination is the lux.

Illumination depends on various factors such as the strength of the light source and the distance of the light source from the object. Light sources closer to an object will typically provide more illumination compared to ones that are farther away. This is because light spreads out as it travels, and thus, the intensity decreases with distance. Understanding illumination helps in settings where the amount of light provided must be controlled carefully, such as in photography or when setting up lighting in a room.

In the context of a photometer setup, understanding how illumination works is crucial to ensure that both sides of the photometer are equally lit. This maintains balance and consistency in measurements or experiments being conducted.
Reflection
Reflection is a fundamental concept in optics where light bounces back when it encounters a surface. When a light ray hits a reflective surface, such as a mirror, it bounces back depending on the angle of incidence and the angle of reflection. These angles are measured from the normal, which is an imaginary line perpendicular to the surface at the point of contact.

In this exercise, the mirror placed near lamp A is perfectly reflective. This means it redirect's A's light back towards the photometer without losing any of its original intensity. This adjustment changes the effective distance that light from A travels to reach the photometer, which is an important consideration in photometric experiments and exercises.

Reflecting surfaces in optical setups can be used to manipulate the path and intensity of light, allowing scientists and engineers to explore various configurations and achieve the desired illumination conditions.
Optics
Optics is the branch of physics that deals with the behavior and properties of light. It encompasses phenomena such as reflection, refraction, diffraction, and interference. Understanding optics is essential for designing experiments and tools that manipulate light, such as lenses, mirrors, and prisms.

In our exercise, the use of a mirror is one of the simplest forms of applying optics. Mirrors help redirect light paths and can be used to adjust how light interacts with objects in their vicinity. Knowledge in optics allows us to calculate and predict how light will behave in various setups, such as the one involving the photometer and lamps A and B.

Optics plays a vital role in everyday technology, including cameras, glasses, and even fiber optic cables that are pivotal in internet data transmission.
Light Distance Calculation
The key to solving the photometry exercise is accurate light distance calculation. This involves understanding how different factors alter the path and effective distance of light from its source to the target.

In this case, with a mirror introduced to lamp A's setup, the effective light path has been extended. Originally the lamp was 60 cm away from the photometer, but the mirror adds an extra 40 cm by reflecting light. This results in a total effective distance of 100 cm.

For both lamps A and B, having equal light distances to the photometer is essential to achieve equal illumination on both sides. Calculating these distances requires careful consideration of all influencing factors, such as mirrored paths or other environmental elements.

Correct light distance calculation ensures precision in any application relying on light distribution, from simple experimental setups to complex industrial processes.

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Most popular questions from this chapter

The power of three sources \(A, B\) and \(C\) are same. The wavelengths emitted by sources are \(4300 \AA, 5550 \AA\) and \(7000 \AA\) respectively. The brightness of sources on the basis of sensation of eye are \(L_{1}, L_{2}\) and \(L_{3}\) respectively. Then: (a) \(L_{1}>L_{2}>L_{3}\) (b) \(L_{1}>L_{3}>L_{2}\) (c) \(L_{2}>L_{1}\) and \(L_{2}>L_{3}\) (d) \(L_{2}

If wavelength \(\lambda_{R}=6200 \AA\) and \(\lambda_{G}=5400 \AA\), then how fast one must rove to see red light signal as a green one? (a) \(4.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (b) \(5.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (c) \(6.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\) (d) \(8.1 \times 10^{7} \mathrm{~m} / \mathrm{s}\)

The intensity produced by a point light source at a small distance \(r\) from the source is proportional to: (a) \(\frac{1}{r}\) (b) \(r^{2}\) (c) \(\frac{1}{r^{2}}\) (d) \(\frac{1}{r^{3}}\)

What must be the velocity of star, if the spectral line \(\lambda=600 \mathrm{~nm}\) of a star shifts by \(0.1 \AA\) towards longer wavelength from the position of the same line in terrestrial laboratory? (Assume the shift to be due to Doppler effect) (a) \(6 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (b) \(5 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (c) \(0.5 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (d) \(6.25 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

The luminous efficiency of the lamp, if luminous flux of 200 watt lamp is 400 lumen is: (a) 2 lumen/watt. (b) 4 lumen/watt (c) 3 lumen/watt (d) 8 lumen/watt

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