Question 1

An object is placed 45.0 cm to the right of a concave mirror with focal length 24.0 cm and -5.0 cm to the left of a lens with focal length -16.0 cm.Where is the image formed by both the mirror and the lens?

Accepted Answer

The answer of An object is placed 45.0 cm to...

Question 2

A concave mirror has radius R.When an object is located a distance 2R from the lens,which describes the image formed?

Accepted Answer

A concave mirror will form a real, inverted image if the object is located beyond the focal point (2R in this case). The image will also be smaller than the object, meaning it is diminished. Therefore, choice A is the best answer.

Question 3

An image is formed using a convex lens,the image being 15 cm past the lens.A second lens is placed 25 cm past the first lens and another image is formed,this time 10 cm past the second lens.Which of the following statements is true?

Accepted Answer

The answer of An image is formed using a convex...

Question 4

A 80-cm focal length thin lens is placed in contact with one of 59.5 cm focal length.A 2.3 cm tall object is placed 55 cm in front of the combination.What is the size of the image?

Accepted Answer

The image formed by the first lens is the object for the second lens. The image formed by the first lens is virtual and 2.3 cm tall. The second lens forms a real image of this virtual image. The magnification of the second lens is -3.7, so the final image is 3.7 cm tall.

Question 5

Use of a parabolic mirror,instead of one made of a circular arc surface,can be used to reduce the occurrence of which of the following effects?

Accepted Answer

Parabolic mirrors are designed to eliminate or minimize spherical aberrations, which occur when the reflected light does not converge on a single point, resulting in blurred images.

Question 6

Five thin lenses,each of focal length f,are placed in contact.What is the resulting focal length of the combination?

Accepted Answer

When thin lenses are placed in contact, their powers add up. The power of a lens is the reciprocal of its focal length (1/f). For five lenses, the total power is 5/f, so the resulting focal length is the reciprocal of this total power, which is f/5.

Question 7

A lens of 10.0-cm focal length is placed 29.0 cm to the right of the object and a mirror of focal length -20.0 cm is placed 49.0 cm to the right of the lens.Where is the image formed by the mirror?

Accepted Answer

Using the lens formula, we can find the position of the object as:
1/f = 1/do - 1/di (where f is the focal length of the lens, do is the distance of the object from the lens, and di is the distance of the image from the lens)
1/10 = 1/29 - 1/di
di = 18.3 cm to the right of the lens

Now, using the mirror formula, we can find the position of the final image as:
1/f_mirror = 1/di - 1/do_mirror (where f_mirror is the focal length of the mirror, and do_mirror is the distance of the object from the mirror)
1/-20 = 1/18.3 - 1/do_mirror
do_mirror = -12.62 cm (the negative sign indicates that the object is behind the mirror)

Therefore, the final image is formed 12.62 cm to the right of the mirror. Answer choice D is the closest to this value.

Question 8

Reducing the lens aperture size is a scheme one can use to reduce the occurrence of which of the following effects?

Accepted Answer

Reducing the lens aperture size is a scheme to reduce the occurrence of spherical aberration.

Question 9

An eyeglass lens is cut with one surface having a radius R₁ and the other surface having a radius R₂,with R₁ < R₂.Both positive radii are measured from the same side with R₁ being the side closer to the eye.Is this convex-concave lens a converging lens or a diverging lens?

Accepted Answer

The answer of An eyeglass lens is cut with one...

Question 10

An image is formed using a convex lens,the image being 15 cm past the lens.A second lens is placed 25 cm past the first lens and another image is formed,this time 10 cm past the second lens.Which of the following statements is always true?

Accepted Answer

Convex lenses are converging lenses, which means they have positive focal lengths. The formation of real images past each lens indicates that both lenses are converging, thus having positive focal lengths.

Question 11

A concave mirror with focal length 17.0 cm is placed 38.0 cm to the left of the object.A convex lens of focal length 18.0 cm is placed 54.0 cm to the right of the object.Where is the image formed by both the mirror and the lens?

Accepted Answer

Using the mirror formula, we have:
1/f = 1/do + 1/di
1/17 = 1/38 + 1/di
di = 25.86 cm (image distance from the mirror)
Using the lens formula, we have:
1/f = 1/di' + 1/do'
1/18 = 1/di' + 1/54
di' = 25.5 cm (image distance from the lens)
The final image distance is the distance between the mirror and the lens:
di_total = di - d_mirror_to_lens - di'
di_total = 25.86 - 16.0 - 25.5
di_total = 13.36 cm
Therefore, the image is formed 13.36 cm to the right of the object, which is choice B.

Question 12

A lens of 13.0-cm focal length is placed 29.0 cm to the right of the object and a mirror of focal length -19.0 cm is placed 39.6 cm to the right of the lens.Where is the final image formed?

Accepted Answer

The answer of A lens of 13.0-cm focal length is...

Question 13

A thin lens has a focal length of 10.00 cm for red light.If the index of refraction for the lens material tends to decrease with increasing wavelength,what is the focal length of the lens for blue light?

Accepted Answer

The focal length of a lens for blue light will be shorter than that for red light if the index of refraction of the lens material decreases with increasing wavelength. This is because blue light has a shorter wavelength than red light, so it is refracted more than red light as it passes through the lens. The lens material with decreasing index of refraction will refract blue light more than it does red light, causing the focal length to be shorter for blue light. Therefore, the focal length for blue light will be less than 10.00 cm.

Question 14

In an ideal case rays coming from an object toward a lens or mirror should be reasonably close to the optic axis.To the extent that this condition is not completely met,which one of the following effects occurs?

Accepted Answer

Spherical aberration occurs when rays from the object are not all focused at the same point due to the curvature of the lens or mirror. This can happen when rays are not close enough to the optic axis. Mirages are an optical phenomenon caused by refraction and do not relate to the given scenario. Chromatic aberration occurs when different wavelengths of light are focused at different points and can occur regardless of the position of rays relative to the optic axis. Light scattering refers to the deviation of light from its original path due to interaction with suspended particles in a material and is not relevant to this scenario.

Question 15

An object is placed 28 cm to the left of a lens of focal length 15 cm.24 cm to the right of this lens is a plane mirror.Where does the final image form?

Accepted Answer

First, we can use the lens formula 1/f = 1/v - 1/u where f = 15 cm, u = -28 cm (since the object is to the left of the lens) to find the image distance v = -11.67 cm. This negative value means that the image is formed to the left of the lens.
Next, we use the mirror formula 1/v' + 1/u' = 1/f where v' is the distance of the final image from the mirror and u' is the distance of the virtual object from the mirror. Since the light from the lens will reflect off the mirror and form a virtual object 24 cm to the left of the mirror, u' = -24 cm. f is still 15 cm, and we can solve for v' to get v' = -8 cm. This negative value means that the final image is formed 8 cm to the left of the mirror. However, since the mirror is to the right of the lens, we need to consider the sign convention for distances: distances to the right of the mirror are positive, while distances to the left of the mirror are negative. Therefore, the final image is actually formed 8 cm to the right of the mirror, which is answer choice B.

Question 16

A fused combination of a diverging and converging lens pair each made from a different index of refraction glass,is used to reduce the occurrence of which of the following effects?

Accepted Answer

A fused combination of a diverging and converging lens pair with different indexes of refraction is called an achromatic lens, which is designed to reduce chromatic aberration. Chromatic aberration is the effect where different wavelengths of light refract differently, causing a blur or color fringing in images. The use of an achromatic lens helps to overcome this problem by bringing the different wavelengths of light to a common focus.

Question 17

A convex lens has a focal length of magnitude F.At which of the following distances from this lens would a real object give an inverted virtual image?

Accepted Answer

A convex lens cannot produce an inverted virtual image; it produces real, inverted images or virtual, upright images depending on the object's position relative to the focal length.

Question 18

A lens has a focal length of 41 cm in air.If this lens were immersed in water,what focal length would result? The index of refraction of the lens is 2.200 and that of water is 1.333.The focal length of a thin lens with index n₁ submerged in a liquid with index n₂ is given by the modified lens maker's formula: )

Accepted Answer

The answer of A lens has a focal length of...

Question 19

Which of the following effects is the result of the fact that the index of refraction of glass will vary with wavelength?

Accepted Answer

Chromatic aberration is the effect that occurs when different wavelengths of light are refracted differently as they pass through a lens or other optical element, resulting in color fringing and blurring of the image. This is caused by the fact that the index of refraction of glass (or other materials) varies with wavelength, with shorter wavelengths being refracted more than longer ones.

Question 20

A 80-cm focal length thin lens is placed in contact with one of 59.2 cm focal length.An object is placed 51 cm in front of the combination.What is the image distance?

Accepted Answer

We can use the lens formula 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance and u is the object distance.
For the given combination of lenses, the effective focal length (feff) can be calculated using the formula 1/feff = 1/f1 + 1/f2, where f1 and f2 are the focal lengths of the individual lenses.
1/feff = 1/80 + 1/59.2
feff ≈ 35.29 cm

Now, using the effective focal length and the object distance, we can find the image distance:
1/feff = 1/v - 1/u
1/35.29 = 1/v - 1/51
v ≈ 102 cm

Therefore, the image distance is 102 cm, which corresponds to choice D.

Quiz 23: Mirrors and Lenses