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A microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.10 mm. Find the index of refraction of the second plate.

Short Answer

Expert verified
The index of refraction of the second plate is approximately 2.69.

Step by step solution

01

Identify the Distance Traveled by Light in the Glass

The light travels through the glass from the bottom surface of the second plate to its upper surface. Between these surfaces, the total movement of the microscope is the 2.10 mm to focus on the upper surface of the second plate.
02

Calculate the Actual Thickness of the Glass Plate

When focusing from the bottom surface to the upper surface of the second plate, the actual thickness that the light perceives is different due to refraction. This thickness that the microscope moves (2.10 mm) is effectively reduced by the refraction index when traveling through the plate.
03

Identify Equations for Index of Refraction (n)

The index of refraction of the second plate (n) can be determined using Snell’s Law or a simplified formula for this scenario: \( n = \frac{d_r}{d_o} \), where \(d_r\) is the distance moved within the medium, and \(d_o\) is the optical path length without the medium. Here, the actual thickness \(d_r\) is 0.780 mm, and the optical path length is 2.10 mm.
04

Calculate Index of Refraction

Use the formula \( n = \frac{d_o}{d_r} \). Substitute the known values: \( n = \frac{2.10}{0.780} \). Calculate to yield \( n \approx 2.69 \).

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Key Concepts

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

Snell's Law
Snell's Law is an essential principle in optics, explaining how light bends when it passes from one medium to another. This bending is known as refraction. The law states that the ratio of the sine of the angle of incidence (the angle at which light hits a surface) to the sine of the angle of refraction (the angle at which it exits) is constant, and this constant is called the index of refraction. It is mathematically expressed as:
  • \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
Where \( n_1 \) and \( n_2 \) are the indices of refraction for the two mediums, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.

In the context of the exercise, while Snell's Law helps us understand how light behaves at the boundaries of different mediums, the formula simplifies further as we assume normal incidence in many cases like straight line travel through glass.Thus, the index of refraction could also be found by understanding the ratios of perceived distances in the medium and vacuum (or air). This understanding allows us to measure the bending effect in simpler terms using the distance the microscope focuses.
Microscope Optics
Microscopes play a crucial role in magnifying tiny objects and usually work using lenses to bend light. However, when a microscope needs to focus through glass, as shown in the exercise, the refraction becomes pivotal.

This exercise describes how light travels through a layered medium. The microscope first focuses on the upper surface of a glass plate via adjusting its optics. When a second plate is placed atop the first, it alters the optical path because the light has to pass through more glass material. Adjusting the microscope further helps it refocus beyond this new boundary. The distance thus adjusted (0.780 mm from the initial 2.10 mm) informs us of how light shifts focus through the second plate.
  • The distance moved by the microscope is less than the actual path light travels in the glass.
  • The calculation uses the perceived distance to find the true distance light travels within the glass or its effective optical path length.
Understanding these optics aids in perceiving how a microscope can adjust to focus through materials with different refractive properties.
Optical Path Length
The concept of Optical Path Length (OPL) is important for accurately describing how light propagates through materials. OPL is a way of measuring the effective distance light travels, accounting for any changes in speed due to the medium. This is particularly relevant when light passes through substances like glass, which alters both its speed and path.

In simple terms, the OPL is the distance light would have traveled in a vacuum instead of the medium it actually traverses. It takes into account the medium's refractive index, using the relation:
  • \( ext{OPL} = n \times d \)
where \( n \) is the index of refraction and \( d \) is the physical distance.

In the given problem, the OPL is the distance the microscope lens had to adjust to shift focus, totaling 2.10 mm when measuring the perceived optical path through the glass. The refractive index helps translate this optical path into understanding how thick the material truly is and how significantly it altered the light's trajectory. Thus, the OPL gives insights into both the medium's properties and the adjustments needed for optical devices to maintain clear images.

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Most popular questions from this chapter

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

A person with a near point of \(85 \mathrm{~cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

A transparent liquid fills a cylindrical tank to a depth of 3.60 m. There is air above the liquid. You look at normal incidence at a small pebble at the bottom of the tank. The apparent depth of the pebble below the liquid's surface is 2.45 m. What is the refractive index of this liquid?

A thin lens with a focal length of 6.00 cm is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 cm from the eye, and that the lens is very close to the eye.

(a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)

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