Question:

Use the mirror equation to deduce that: (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the location of the object. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Answer:

(a) Use the mirror equation 1/f = 1/v + 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

Substitute f for 2f, and rearrange the equation to obtain v = 2f(u-f)/(u-2f).

For a real image, v > 0. Thus, 2f(u-f) > 0, which implies u > 2f.

Therefore, an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

(b) Use the mirror equation 1/f = 1/v + 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

For a convex mirror, f is negative. Thus, the equation becomes 1/f = 1/v + 1/u, where f is negative.

Rearrange the equation to obtain v = f(u-f)/(u-f).

For a virtual image, v < 0. Thus, f(u-f) < 0, which implies u < f.

Therefore, a convex mirror always produces a virtual image independent of the location of the object.

(c) Use the mirror equation 1/f = 1/v + 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

For a convex mirror, f is negative. Thus, the equation becomes 1/f = 1/v + 1/u, where f is negative.

Rearrange the equation to obtain v = f(u-f)/(u-f).

For a virtual image, v < 0. Thus, f(u-f) < 0, which implies u > f.

Therefore, the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

(d) Use the mirror equation 1/f = 1/v + 1/u, where f is the focal length, v is the distance of the image from the mirror, and u is the distance of the object from the mirror.

For a concave mirror, f is positive. Thus, the equation becomes 1/f = 1/v + 1/u, where f is positive.

Rearrange the equation to obtain v = f(u-f)/(u-f).

For a virtual image, v < 0. Thus, f(u-f) < 0, which implies u < f.

Therefore, an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Question:

A person is looking at another person wearing a shirt with a pattern comprising of vertical and horizontal lines. He is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Answer:

1. This defect is due to astigmatism, which is a condition in which the eye’s cornea is not evenly curved.

2. Astigmatism can be corrected by wearing eyeglasses or contact lenses that have a special curvature to correct the unevenness of the cornea. Laser eye surgery is also an option to correct astigmatism.

Question:

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?

Answer:

(a) Convex Lens of Focal Length 20 cm:

1. The beam of light converges at point P.
2. The lens is placed 12 cm from point P.
3. The focal length of the lens is 20 cm.
4. The beam of light will now converge at a point 8 cm from the lens (20 cm - 12 cm = 8 cm).

Question:

Answer the following questions: (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain. (b) A virtual image, we always say, cannot be caught on a screen. Yet when we see a virtual image, we are obviously bringing it on to the screen (i.e., the retina) of our eye. Is there a contradiction? (c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is? (d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease? (e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?

Answer:

(a) Yes, plane and convex mirrors can produce real images under certain circumstances. For example, when a convex mirror is used to form a real image, the object must be placed at a certain distance from the mirror.

(b) No, there is no contradiction. The virtual image is formed in the eye, but it cannot be projected onto a screen.

(c) The fisherman would look shorter to the diver than what he actually is.

(d) Yes, the apparent depth of a tank of water does change when viewed obliquely. The apparent depth will decrease when viewed obliquely.

(e) Yes, this fact is of great use to a diamond cutter. The high refractive index of diamond helps in cutting the diamond into desired shapes and sizes.

Question:

A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens.

Answer:

1. Draw a diagram of the setup, labeling the distances and noting the object, screen, and lens.

2. Calculate the distance between the object and the lens: 90 cm - 20 cm = 70 cm.

3. Calculate the distance between the lens and the screen: 90 cm + 20 cm = 110 cm.

4. Use the lens equation to calculate the focal length (f): 1/f = (1/d1) + (1/d2), where d1 is the distance between the object and the lens and d2 is the distance between the lens and the screen.

5. Substitute the values for d1 and d2 into the equation and solve for f: 1/f = (1/70) + (1/110) => f = 55 cm.

The focal length of the lens is 55 cm.

Question:

A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?

Answer:

Answer:

1. The distance by which the pin appears to be raised is equal to the thickness of the glass slab, which is 15 cm.
2. No, the answer does not depend on the location of the slab. The refractive index of the glass is constant, so the distance by which the pin appears to be raised will remain the same.

Question:

(a) Determine the effective focal length of the combination of the two lenses, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40cm. Determine the magnification produced by the two-lens system, and the size of the image.

Answer:

(a) The effective focal length of the combination of the two lenses is equal to the sum of the focal lengths of the individual lenses. The answer does not depend on which side of the combination a beam of parallel light is incident. The notion of effective focal length of this system is useful in determining the power of the combination of lenses.

(b) The magnification produced by the two-lens system is equal to the ratio of the image distance to the object distance, or M = -d_image/d_object. The image distance can be calculated using the lens equation: 1/d_image + 1/d_object = 1/f, where f is the effective focal length of the two lenses. Substituting the given values, we get 1/d_image + 1/40 = 1/f, which gives d_image = -40f/ (f-40). The size of the image is equal to the magnification multiplied by the size of the object, or s_image = M*s_object. Substituting the values, we get s_image = -40f/ (f-40)*1.5 cm.

Question:

A card sheet divided into squares each of size 1mm2 is being viewed at a distance of 9cm through a magnifying glass (a converging lens of focal length 9cm) held close to the eye. In order to view the squares distinctly with the maximum possible magnifying power. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain:

Answer:

(a) The magnification produced by the lens is 1, as the focal length of the lens is equal to the distance of the object from the lens. The area of each square in the virtual image is also 1mm2.

(b) The angular magnification (magnifying power) of the lens is 1, as the angular size of the object is the same as the angular size of the image.

(c) Yes, the magnification in (a) is equal to the magnifying power in (b). This is because the angular magnification is the ratio of the angular size of the image to the angular size of the object, which in this case is 1:1.

Question:

Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Answer:

1. Short-sightedness (myopia) and long-sightedness (hypermetropia) do not necessarily imply that the eye has partially lost its ability of accommodation.

2. These defects of vision can be caused by the shape of the eyeball being too long or too short, which affects the way light is refracted, resulting in blurred vision. It can also be caused by the cornea or lens not being curved correctly, or by the eye muscles not working properly.

Question:

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Answer:

1. First, calculate the distance from the lens to the image. This distance is equal to the sum of the distance from the lens to the bulb (3m) and the distance from the bulb to the image (3m). Therefore, the distance from the lens to the image is 6m.

2. Next, calculate the focal length of the lens. The focal length of a convex lens is equal to the inverse of the sum of the inverse of the object distance (1/3m) and the inverse of the image distance (1/6m). Therefore, the focal length of the lens is 2m.

3. Finally, calculate the maximum possible focal length of the lens. The maximum possible focal length of the lens is equal to twice the focal length of the lens, which is 4m. Therefore, the maximum possible focal length of the lens required for the purpose is 4m.

Question:

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.

Answer:

1. Separation between the two lenses = 25 cm - (8.0 mm + 2.5 cm) = 14.5 cm

2. Magnifying power of the microscope = (Object distance/Focal length of objective) x (Focal length of eyepiece/Near point) = (9.0 mm/8.0 mm) x (2.5 cm/25 cm) = 0.1125 x 0.1 = 0.01125 = 11.25

Question:

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Answer:

1. Identify the given information: Refractive index = 1.55 Focal length = 20 cm

2. Calculate the radius of curvature: Use the equation 1/f = (n-1) (1/R1 + 1/R2)

1/20 = (1.55 - 1) (1/R1 + 1/R2)

R1 = R2 = 20 cm / (1.55 - 1) = 25.8 cm

Question:

A card sheet divided into squares each of size 1mm2 is being viewed at a distance of 9cm through a magnifying glass (a converging lens of focal length 9cm) held close to the eye and the magnifying glass if the virtual image of each square is to have an area of 6.25mm^2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

Answer:

1. Calculate the magnification of the magnifying glass: m = (9 cm / 1 mm) = 900

2. Calculate the size of the virtual image of each square: s = (6.25 mm^2) / (900) = 0.0069 mm

3. Determine if the virtual image would be visible to the eye: Yes, the virtual image of each square would be visible to the eye because it is larger than the resolution of the human eye (which is about 0.1 mm).

Question:

A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a)the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25cm)?

Answer:

(a) Magnifying power of the telescope = Focal length of the objective lens / Focal length of the eyepiece = 140cm / 5.0cm = 28

(b) Magnifying power of the telescope = Focal length of the objective lens / (Focal length of the eyepiece - Distance of the final image from the eyepiece) = 140cm / (5.0cm - 25cm) = 140cm / (-20cm) = -7

Question:

Answer the following questions: (a) For the telescope described in Exercise (a)[A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?],
What is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100m tall tower 3km away, what is the height of the image of the tower formed
by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25cm?

Answer:

(a) The separation between the objective lens and the eyepiece is 135 cm.

(b) The height of the image of the tower formed by the objective lens is 0.75 m.

(c) The height of the final image of the tower if it is formed at 25 cm is 6.25 cm.

Question:

An object of size 3.0 cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

Answer:

1. The image produced by the lens is a real, inverted image that is 7 cm in size.

2. If the object is moved further away from the lens, the image will become smaller and less sharp. The image will also be further away from the lens.

Question:

You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion, (b) disperse (and displace) a pencil of white light without much deviation.

Answer:

A) To deviate a pencil of white light without much dispersion:

1. Choose a crown glass prism with a large angle of refraction (e.g. 60°).
2. Place the prism at an angle to the light beam so that the light is refracted at the angle of refraction.
3. Place a second prism (made of either crown glass or flint glass) behind the first prism, with its angle of refraction facing the same direction as the first prism.
4. The two prisms should be placed close together so that the light is refracted at both angles of refraction, thus deviating the light.

B) To disperse (and displace) a pencil of white light without much deviation:

1. Choose a flint glass prism with a small angle of refraction (e.g. 30°).
2. Place the prism at an angle to the light beam so that the light is refracted at the angle of refraction.
3. Place a second prism (made of either crown glass or flint glass) behind the first prism, with its angle of refraction facing the opposite direction of the first prism.
4. The two prisms should be placed close together so that the light is refracted at both angles of refraction, thus dispersing the light and displacing it in the opposite direction.

Question:

Answer the following questions: (a )The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass, one usually positions ones eyes very close to the lens. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?

Answer:

(a) A magnifying glass provides angular magnification because it increases the angle subtended at the eye by an object.

(b) Yes, angular magnification will change if the eye is moved back. It will decrease as the eye moves further away from the lens.

(c) We cannot use a convex lens of smaller and smaller focal length to achieve greater magnifying power because the lens would become too thin and weak to be practical.

(d) Both the objective and the eyepiece of a compound microscope must have short focal lengths to ensure that the image is magnified enough to be visible to the eye.

(e) Our eyes should be positioned a short distance away from the eyepiece for best viewing because if the eye is too close to the eyepiece, the image will be blurred. The short distance should be about 10-20 cm.

Question:

For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye - lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye. A 10 to 14 D B 20 to 24 D C 28 to 32 D D 14 to 18 D

Answer:

Answer: B 20 to 24 D

Question:

What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

Answer:

1. The focal length of the combined system is 10 cm.
2. The system is a converging lens.

Question:

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40o. What is the refractive index of the material of the prism? The refracting angle of the prism is 60o. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Answer:

1. The refractive index of the material of the prism can be calculated using the formula: n = sin(θm/2)/sin(θi/2) where θm is the angle of minimum deviation and θi is the refracting angle of the prism.

2. Therefore, n = sin(40o/2)/sin(60o/2) = 0.8

3. If the prism is placed in water (refractive index 1.33), the new angle of minimum deviation of a parallel beam of light can be calculated using the formula: θm = 2 x arcsin(n x sin(θi/2)) where n is the refractive index of the medium and θi is the refracting angle of the prism.

4. Therefore, θm = 2 x arcsin(1.33 x sin(60o/2)) = 48.1o

Question:

A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm. (a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass? (b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?

Answer:

a) The closest distance at which the man should keep the lens from the page is 5 cm (the focal length of the lens) and the farthest distance at which he should keep the lens from the page is 25 cm (the normal near point).

b) The maximum angular magnification (magnifying power) possible using the above simple microscope is 5 (the focal length of the lens divided by the normal near point) and the minimum angular magnification (magnifying power) possible using the above simple microscope is 1 (the normal near point divided by the focal length of the lens).

Question:

A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

Answer:

Step 1: Calculate the radius of the water surface through which light can emerge out. This can be done using the formula:

Radius = (Refractive index of water - 1) * Depth of water

Radius = (1.33 - 1) * 80 cm Radius = 80 cm

Step 2: Calculate the area of the water surface through which light can emerge out. This can be done using the formula:

Area = π * Radius2

Area = π * (80 cm)2 Area = 6,400 cm2

Question:

A small candle, 2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Answer:

1. To obtain a sharp image, the screen should be placed at a distance of 54 cm from the mirror.

2. The nature and size of the image will be an inverted and magnified virtual image of the candle.

3. If the candle is moved closer to the mirror, the screen should be moved closer to the mirror as well, in order to maintain the same distance between the image and the screen.

Question:

(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48×10^6m, and the radius of lunar orbit is 3.8×10^8m.

Answer:

(a) Angular magnification = (focal length of objective lens)/(focal length of eyepiece) = 15m/0.01m = 1500

(b) The angular diameter of the moon = (diameter of moon)/(radius of lunar orbit) = 3.48x10^6m/3.8x10^8m = 0.00914 radians

The diameter of the image of the moon formed by the objective lens = (angular diameter of the moon) x (angular magnification of the telescope) = 0.00914 radians x 1500 = 13.71m

Question:

A small telescope has an objective lens of focal length 144cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?

Answer:

1. Magnifying power of the telescope = 144 cm/6 cm = 24

2. Separation between the objective and the eyepiece = 144 cm - 6 cm = 138 cm

Question:

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Answer:

1. The location of the image is 3 cm away from the mirror, on the same side as the needle.

2. The magnification is -3, meaning the image is inverted and three times the size of the object.

3. As the needle is moved farther away from the mirror, the location of the image moves farther away from the mirror as well, and the magnification decreases.

Question:

A myopic person has been using spectacles of power −1.0 D for distant vision. During old age, he also needs to use a separate reading glass of power +2.0 D. Explain what may have happened.

Answer:

1. A myopic person has a refractive error in their eyes, meaning that the light entering their eyes is not focused properly, resulting in blurred vision.

2. In order to correct this refractive error, they may have been prescribed spectacles of power −1.0 D, which is a measure of the lens strength that is needed to correct the refractive error.

3. As the person gets older, the lens of the eye begins to harden, which can cause further refractive errors. In this case, the person may have developed a condition known as presbyopia, which is a condition that causes difficulty with near vision.

4. To correct this new refractive error, the person may be prescribed a separate reading glass of power +2.0 D, which is a measure of the lens strength that is needed to correct the presbyopia.

Question:

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at: (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Answer:

(a) To obtain the final image at the least distance of distinct vision (25cm), the object should be placed at a distance of 15.25cm from the objective lens.

(b) To obtain the final image at infinity, the object should be placed at a distance of 22.25cm from the objective lens.

(a) The magnifying power of the microscope at the least distance of distinct vision (25cm) is 3.91.

(b) The magnifying power of the microscope at infinity is 4.18.

Question:

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

Answer:

1. Refractive index of water = (12.5 cm - 9.4 cm)/9.4 cm = 0.31

2. The distance the microscope would have to be moved to focus on the needle again would be (12.5 cm - 9.4 cm) x (1.63 - 0.31)/0.31 = 4.7 cm.

Question:

An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25cm and an eyepiece of focal length 5cm. How will you set up the compound microscope?

Answer:

1. Place the objective lens on the microscope body, with the distance between the lens and the stage adjusted to 1.25 cm.

2. Place the eyepiece lens in the ocular tube, with the distance between the lens and the objective lens adjusted to 5 cm.

3. Adjust the focus knob until the image is in focus.

4. Adjust the interpupillary distance of the ocular tubes until the image is clear.

5. Adjust the diopter of the eyepiece until the angular magnification of 30X is achieved.

Question:

A card sheet divided into squares each of size 1mm2 is being viewed at a distance of 9cm through a magnifying glass (a converging lens of focal length 9cm) held close to the eye.(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?(b) What is the angular magnification (magnifying power) of the lens?(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain

Answer:

a) The magnification produced by the lens is equal to the ratio of the image distance to the object distance. In this case, the image distance is 0 cm (since the lens is held close to the eye) and the object distance is 9 cm. Therefore, the magnification is 0/9 = 0. The area of each square in the virtual image is 0 mm2.

b) The angular magnification (magnifying power) of the lens is equal to the ratio of the angle subtended by the image to the angle subtended by the object. In this case, the angle subtended by the image is 0° and the angle subtended by the object is 9°. Therefore, the angular magnification (magnifying power) of the lens is 0/9 = 0.

c) No, the magnification in (a) is not equal to the magnifying power in (b). This is because magnification refers to the ratio of the size of the image to the size of the object, while magnifying power refers to the ratio of the angle subtended by the image to the angle subtended by the object.

Question:

At what angle should a ray of light be incident on the face of a prism of refracting angle 60∘ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.

Answer:

1. The angle of incidence should be greater than the refracting angle of the prism, i.e. 60°.

2. The angle of incidence should be equal to the critical angle for the material of the prism, which is equal to the inverse of the refractive index of the material.

3. Therefore, the angle of incidence should be equal to 1/1.524 = 65.6°.