Question

Light from a very tiny light bulb passes through a hole of diameter $0.15 \mathrm{m}$ in a dark screen that is $2.5 \mathrm{m}$ away (Fig. $\mathrm{P} 24.1$ ). Assume the light travels as ideal rays as it passes from the bulb, through the hole, and onto a second screen on the far right in Figure $P 24.1 .$ If the light rays illuminate a spot of diameter $0.65 \mathrm{m}$ on the second screen, what is the spacing between the two screens?

Name: Problem 1, Chapter 24: College Physics: Reasoning and Relationships
Uploaded: 2022-02-14T04:12:32-08:00
Duration: 1 min 5 s
Channel: Raj Bala
Description: Light from a very tiny light bulb passes through a hole of diameter 0.15 m in a dark screen that is 2.5 m away (Fig. P 24.1 ). Assume the light travels as ideal rays as it passes from the bulb, through the hole, and onto a second screen on the far right in Figure P 24.1 . If the light rays illuminate a spot of diameter 0.65 m on the second screen, what is the spacing between the two screens?

   Light from a very tiny light bulb passes through a hole of diameter $0.15 \mathrm{m}$ in a dark screen that  is $2.5 \mathrm{m}$ away (Fig. $\mathrm{P} 24.1$ ). Assume the light travels as ideal rays as it passes from the bulb, through the hole, and onto a second screen on the far right in Figure $P 24.1 .$ If the light rays illuminate a spot of diameter $0.65 \mathrm{m}$ on the second screen, what is the spacing between the two screens?

College Physics: Reasoning and Relationships

Nicholas J.… 2nd Edition

Chapter 24, Problem 1

Instant Answer

Solved by Expert Raj Bala

02/14/2022

Step 1

We know that $\tan(\theta) = \frac{E}{S}$, which can be rewritten as $\tan(\theta) = \frac{Y_1}{X}$, where $Y_1$ is the radius of the hole in the screen and $X$ is the distance from the light bulb to the screen. Since the radius is half the diameter, we can Show more…

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Light from a very tiny light bulb passes through a hole of diameter $0.15 \mathrm{m}$ in a dark screen that is $2.5 \mathrm{m}$ away (Fig. $\mathrm{P} 24.1$ ). Assume the light travels as ideal rays as it passes from the bulb, through the hole, and onto a second screen on the far right in Figure $P 24.1 .$ If the light rays illuminate a spot of diameter $0.65 \mathrm{m}$ on the second screen, what is the spacing between the two screens?

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Key Concepts

Geometrical Optics

This principle assumes that light travels in straight lines and can be represented by rays. In problems like this, it allows one to model how light propagates from a source, passes through an aperture, and then forms an image on a screen without considering effects like diffraction, making it easier to analyze the system using simple geometric relationships.

Similar Triangles

This concept is fundamental in relating the sizes of objects and their images when projected through an aperture. By forming triangles between the light source, the edges of the hole, and the illuminated spot on the screen, one can use the proportionality of sides to determine missing distances or dimensions in the system.

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