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Chapter 31: Problem 1

How can you see a virtual image, when it's not "really there"?

Short Answer

Expert verified
We can see a virtual image because our eyes perceive the diverging light rays as converging. This makes it seem as if the light is coming 'from' the virtual image, which is actually not physically present. A virtual image is just an apparent point of intersection of the light rays, formed by the refraction of light in optical devices, such as lenses or mirrors.

Step by step solution

01

Understanding Virtual Images

A virtual image is the image formed by the apparent intersection of the light rays from an object, which would converge if extended back on the same side as the object. They are formed by the refraction of light in devices such as lenses and mirrors. The image isn't 'really there' because you can't project it on a screen as it can be with a real image. The light rays are diverging, making the image appear to be originated from a specific location if traced back.
02

How Light Rays Work

The light rays from the object diverge after passing through or reflecting off the optical device (lens or mirror), but our eyes see them as converging. The image appears upright and located behind the device.
03

Perception by Our Eyes

Our eyes perceive the diverging light rays as if they are converging from the position where the virtual image appears. Hence, our eyes interpret the light as if it's coming 'from' the image rather than the actual object. This gives the illusion of a virtual image being 'there' even if it's not physically present.

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Most popular questions from this chapter

Under what circumstances will the image in a concave mirror be the same size as the object?

An object \(15 \mathrm{cm}\) from a concave mirror has a virtual image magnified 2.5 times. What's the mirror's focal length?

Generalize the derivation of the lensmaker's formula (Equation 31.7 ) to show that a lens of refractive index \(n_{\text {lens }}\) in an external medium with index \(n_{\mathrm{ext}}\) has focal length given by $$\frac{1}{f}=\left(\frac{n_{\mathrm{lens}}}{n_{\mathrm{ext}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$

If you placed a screen at the location of a virtual image, would the image appear on the screen? Why or why not?

By what factor is the image magnified for an object 1.5 focal lengths from a converging lens? Is the image upright or inverted?
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