401. The spectacle lens used to correct myopia is
ⓐ. a convex lens of positive power
ⓑ. a plane glass slab of zero power
ⓒ. a cylindrical lens for every case
ⓓ. a concave lens of negative power
Correct Answer: a concave lens of negative power
Explanation: Myopia is corrected by placing a concave lens in front of the eye. The concave lens diverges incoming rays from a distant object before they enter the eye. This makes the rays appear to come from the far point of the myopic eye. The eye can then focus them on the retina. Since a concave lens has negative focal length in air, its power is negative.
402. A myopic person has a far point at \(2.0\,m\). The spectacle lens needed for viewing distant objects should have power
ⓐ. \(-0.50\,\text{D}\)
ⓑ. \(+0.50\,\text{D}\)
ⓒ. \(-2.0\,\text{D}\)
ⓓ. \(+2.0\,\text{D}\)
Correct Answer: \(-0.50\,\text{D}\)
Explanation: \( \textbf{Given data:} \)
The far point of the myopic eye is \(2.0\,m\).
A distant object is effectively at infinity:
\[
u=-\infty
\]
The spectacle lens must form a virtual image at the far point:
\[
v=-2.0\,m
\]
Use the lens formula:
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Since \(\frac{1}{u}=0\) for an object at infinity,
\[
\frac{1}{f}=\frac{1}{-2.0}-0
\]
\[
f=-2.0\,m
\]
Power is
\[
P=\frac{1}{f}=\frac{1}{-2.0}
\]
\[
P=-0.50\,\text{D}
\]
\( \textbf{Final answer:} \) \(-0.50\,\text{D}\), so the correcting lens is concave.
403. A myopic eye has far point \(50\,cm\). For clear viewing of very distant objects, the required spectacle power is
ⓐ. \(-2.0\,\text{D}\)
ⓑ. \(+2.0\,\text{D}\)
ⓒ. \(-0.50\,\text{D}\)
ⓓ. \(+0.50\,\text{D}\)
Correct Answer: \(-2.0\,\text{D}\)
Explanation: \( \textbf{Known condition:} \)
A distant object has
\[
u=-\infty
\]
The corrective lens should form a virtual image at the far point:
\[
v=-50\,cm=-0.50\,m
\]
Use
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
For \(u=-\infty\),
\[
\frac{1}{u}=0
\]
Therefore,
\[
\frac{1}{f}=\frac{1}{-0.50}
\]
\[
f=-0.50\,m
\]
Power is
\[
P=\frac{1}{f}=-2.0\,\text{D}
\]
\( \textbf{Final answer:} \) \(-2.0\,\text{D}\), with the negative sign showing a diverging correcting lens.
404. A claim says, “A myopic eye should be corrected by a convex lens because a distant object sends parallel rays.” The better response is that a myopic eye needs
ⓐ. a convex lens to increase convergence further
ⓑ. no lens because parallel rays never need focusing
ⓒ. a concave lens to reduce effective convergence
ⓓ. a prism because prisms always form retinal images
Correct Answer: a concave lens to reduce effective convergence
Explanation: A myopic eye already brings rays from distant objects to a focus before the retina. Adding a convex lens would increase convergence and worsen this focusing error. A concave lens spreads the incoming rays slightly before they enter the eye. The myopic eye then focuses these rays on the retina as if they came from its far point. The correction is chosen according to where the eye forms the image, not merely according to the fact that distant rays are parallel.
405. Hypermetropia is a defect in which a person usually has difficulty seeing
ⓐ. distant objects clearly while nearby objects are always clear
ⓑ. nearby objects clearly
ⓒ. only objects at infinity
ⓓ. only objects through a prism
Correct Answer: nearby objects clearly
Explanation: Hypermetropia is also called far-sightedness. A hypermetropic eye can generally see distant objects more easily than nearby objects. For a near object, the image tends to form behind the retina if the eye cannot provide enough convergence. This may happen because the eyeball is too short or the eye lens has insufficient converging power. The correction uses a converging lens to help bring the image onto the retina.
406. The spectacle lens used to correct hypermetropia is
ⓐ. a concave lens of negative power
ⓑ. a convex lens of positive power
ⓒ. a plane mirror
ⓓ. a rectangular glass slab
Correct Answer: a convex lens of positive power
Explanation: A hypermetropic eye does not converge rays from nearby objects enough to form the image on the retina. A convex spectacle lens adds convergence before the rays enter the eye. This helps the eye focus a near object clearly. Since a convex lens has positive focal length in air, its power is positive. The correction is the opposite of myopia, where a concave lens is used.
407. A hypermetropic person has near point \(75\,cm\). What power of spectacle lens is needed so that an object at \(25\,cm\) can be seen clearly?
ⓐ. \(-2.67\,\text{D}\)
ⓑ. \(+1.33\,\text{D}\)
ⓒ. \(+2.67\,\text{D}\)
ⓓ. \(-1.33\,\text{D}\)
Correct Answer: \(+2.67\,\text{D}\)
Explanation: \( \textbf{Required correction:} \)
The object should be viewed at the normal near point:
\[
u=-25\,cm
\]
The corrective lens must form a virtual image at the person's near point:
\[
v=-75\,cm
\]
Use the lens formula:
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Substitute:
\[
\frac{1}{f}=\frac{1}{-75}-\frac{1}{-25}
\]
\[
\frac{1}{f}=-\frac{1}{75}+\frac{1}{25}
\]
\[
\frac{1}{f}=\frac{-1+3}{75}=\frac{2}{75}
\]
\[
f=37.5\,cm=0.375\,m
\]
Power is
\[
P=\frac{1}{0.375}\approx+2.67\,\text{D}
\]
\( \textbf{Final answer:} \) \(+2.67\,\text{D}\), so the correcting lens is convex.
408. A hypermetropic person's near point is \(50\,cm\). The spectacle lens should allow reading at \(25\,cm\). The required power is
ⓐ. \(-2.0\,\text{D}\)
ⓑ. \(+4.0\,\text{D}\)
ⓒ. \(-4.0\,\text{D}\)
ⓓ. \(+2.0\,\text{D}\)
Correct Answer: \(+2.0\,\text{D}\)
Explanation: \( \textbf{Given condition:} \)
The book is at the normal near point:
\[
u=-25\,cm
\]
The lens should form a virtual image at the person's near point:
\[
v=-50\,cm
\]
Lens formula:
\[
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
\]
Substitute:
\[
\frac{1}{f}=\frac{1}{-50}-\frac{1}{-25}
\]
\[
\frac{1}{f}=-\frac{1}{50}+\frac{1}{25}
\]
\[
\frac{1}{f}=\frac{-1+2}{50}=\frac{1}{50}
\]
\[
f=50\,cm=0.50\,m
\]
Power is
\[
P=\frac{1}{0.50}=+2.0\,\text{D}
\]
\( \textbf{Final answer:} \) \(+2.0\,\text{D}\), a positive-power reading correction.
409. A comparison of vision defects is shown below:
| Row | Defect | Main difficulty | Correction |
|---|
| P | Myopia | distant objects | concave lens |
| Q | Hypermetropia | nearby objects | convex lens |
| R | Myopia | image of distant object before retina | diverging lens |
| S | Hypermetropia | image of near object tends behind retina | concave lens |
The acceptable rows are
ⓐ. P, Q, and R only
ⓑ. P and S only
ⓒ. Q, R, and S only
ⓓ. P, Q, R, and S
Correct Answer: P, Q, and R only
Explanation: Row P is acceptable because myopia affects distant vision and is corrected using a concave lens. Row Q is acceptable because hypermetropia affects near vision and is corrected using a convex lens. Row R is acceptable because in myopia the distant-object image forms before the retina, and a diverging lens helps shift the effective focus back. Row S is not acceptable because hypermetropia needs a convex lens, not a concave lens. The lens type must compensate the direction of the focusing error.
410. Presbyopia commonly occurs with age because
ⓐ. the prism angle of the eye becomes zero
ⓑ. the retina begins acting like a plane mirror
ⓒ. the power of accommodation of the eye decreases
ⓓ. the speed of light in the eye becomes infinite
Correct Answer: the power of accommodation of the eye decreases
Explanation: Presbyopia is an age-related defect of vision. With age, the eye lens becomes less flexible and the ciliary muscles become less effective in changing its curvature. As a result, the eye loses some of its power of accommodation. Near objects become harder to focus clearly. This defect is different from ordinary myopia or hypermetropia because it is mainly linked with reduced accommodation.
411. Bifocal lenses are often used when a person needs correction for
ⓐ. only dispersion by a prism
ⓑ. only total internal reflection in the eye
ⓒ. only colour blindness
ⓓ. both distant and near vision
Correct Answer: both distant and near vision
Explanation: A bifocal lens has two different optical powers in different parts of the same spectacle lens. One part is used for distant vision, and the other part is used for near vision. Such lenses are useful for people who have difficulty with both ranges, commonly in presbyopia combined with another refractive defect. The idea is not related to prism dispersion or total internal reflection. The different powers help the eye focus objects at different distances.
412. Astigmatism is commonly corrected using
ⓐ. a plane mirror
ⓑ. a cylindrical lens
ⓒ. a rectangular glass slab
ⓓ. a triangular prism only
Correct Answer: a cylindrical lens
Explanation: Astigmatism occurs when the eye does not focus light equally in all meridians, often due to unequal curvature of the cornea or lens. A cylindrical lens has power in one direction more than another. This helps compensate for the unequal focusing of the eye. A spherical convex or concave lens alone may not fully correct the directional focusing error. The correction is therefore matched to the axis and amount of astigmatism.
413. Consider these statements about the human eye and vision correction:
I. A normal eye has near point approximately \(25\,cm\).
II. Myopia is corrected by a concave lens.
III. Hypermetropia is corrected by a convex lens.
IV. Astigmatism is corrected only by increasing the prism angle of spectacles.
The suitable set is
ⓐ. I and IV only
ⓑ. I, II, and III only
ⓒ. II, III, and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and III only
Explanation: Statement I is suitable because the near point of a normal eye is taken as \(25\,cm\). Statement II is suitable because a concave lens corrects myopia by diverging rays before they enter the eye. Statement III is suitable because a convex lens corrects hypermetropia by adding convergence for near objects. Statement IV is not suitable because astigmatism is corrected using cylindrical lenses, not by simply increasing a prism angle. Each defect needs a correction that matches its optical cause.
414. Assertion: A convex lens is used for correcting hypermetropia.
Reason: The hypermetropic eye needs additional convergence to focus near objects on the retina.
ⓐ. Both are true, and Reason explains Assertion
ⓑ. Both are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both are true, and Reason explains Assertion
Explanation: The assertion is true because hypermetropia is corrected using a convex lens. The reason is also true because a hypermetropic eye does not provide enough convergence for nearby objects. The convex lens supplies additional convergence before the rays enter the eye. This makes the near object appear to be at a distance the eye can focus comfortably. The correction directly addresses the image-forming tendency behind the retina.
415. A simple microscope is basically
ⓐ. a concave lens used to form only diminished images
ⓑ. a plane mirror used to laterally invert an object
ⓒ. a convex lens used to view a nearby small object
ⓓ. a prism used only to disperse white light
Correct Answer: a convex lens used to view a nearby small object
Explanation: A simple microscope is also called a magnifying glass. It uses a convex lens because a convex lens can form a virtual, erect, enlarged image when the object is placed within its focal length. The eye views this enlarged virtual image and sees the object under a larger visual angle. A concave lens gives a diminished virtual image for a real object, so it is not used as a simple magnifier. The device works by refraction through a lens, not by mirror reflection or prism dispersion.
416. For a convex lens to act as a simple microscope, the object should be placed
ⓐ. exactly at \(2F\) on the object side
ⓑ. far beyond \(2F\)
ⓒ. at infinity
ⓓ. between \(O\) and \(F\)
Correct Answer: between \(O\) and \(F\)
Explanation: A convex lens forms a virtual, erect, enlarged image only when the object lies inside the focal length. This means the object is placed between the optical centre \(O\) and the principal focus \(F\). The refracted rays then diverge, and their backward extensions meet on the same side as the object. If the object is outside the focal length, the image becomes real and inverted. A simple microscope needs the enlarged virtual image condition for comfortable direct viewing.
417. The image formed by a simple microscope for normal magnifying use is
ⓐ. real, inverted, and diminished
ⓑ. real, erect, and same size
ⓒ. virtual, inverted, and diminished
ⓓ. virtual, erect, and enlarged
Correct Answer: virtual, erect, and enlarged
Explanation: In a simple microscope, the object is placed within the focal length of a convex lens. The lens then produces diverging refracted rays. The eye traces these rays backward and sees a virtual image on the same side as the object. The image is erect because its height has the same orientation as the object. It is enlarged because the object is close to the focus from inside, so the visual angle at the eye increases.
418. A simple microscope increases apparent size mainly by increasing
ⓐ. the actual height of the object
ⓑ. the visual angle at the eye
ⓒ. the refractive index of air around the eye
ⓓ. the wavelength of light in vacuum
Correct Answer: the visual angle at the eye
Explanation: A magnifying glass does not physically increase the object's actual size. It allows the object to be placed closer to the eye than the normal near point while still being seen clearly through the lens. This increases the visual angle subtended at the eye. A larger visual angle makes the object appear bigger. The magnification is therefore angular in nature, not a mechanical increase in object dimensions.
419. For a simple microscope, the magnifying power when the final image is formed at the near point is
ⓐ. \(M=\frac{D}{f}\)
ⓑ. \(M=\frac{f}{D}\)
ⓒ. \(M=1+\frac{D}{f}\)
ⓓ. \(M=1+\frac{f}{D}\)
Correct Answer: \(M=1+\frac{D}{f}\)
Explanation: When the final image is formed at the near point, the eye views the image at the least distance of distinct vision \(D\). In this condition, the simple microscope gives angular magnification \(M=1+\frac{D}{f}\). The extra \(1\) appears because the final image is at the near point rather than at infinity. Here \(f\) is the focal length of the convex lens and \(D\) is usually taken as \(25\,cm\). A shorter focal length gives larger magnifying power.
420. For a simple microscope used with relaxed eye, the final image is at infinity. Its magnifying power is
ⓐ. \(M=\frac{D}{f}\)
ⓑ. \(M=1+\frac{D}{f}\)
ⓒ. \(M=\frac{f}{D}\)
ⓓ. \(M=D+f\)
Correct Answer: \(M=\frac{D}{f}\)
Explanation: A relaxed eye means the final image is at infinity. For a simple microscope in this condition, the object is placed at the focus of the convex lens. The emergent rays are parallel, so the eye views them without accommodation strain. The angular magnifying power is \(M=\frac{D}{f}\). This is smaller than the near-point value \(1+\frac{D}{f}\) for the same lens.
