UET UG Physics Booster Test 3 - Optical Instruments -

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CUET UG Physics Booster Test 3 - Optical Instruments

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3 - Optical Instruments

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QUESTION 1 OF 20

Consider the statements on simple microscope magnifying lens. Choose the correct statement:
1. The image formed is virtual and closer than infinity if the object is closer than f.
2. It allows the object to be brought closer to the eye than D.
3. Magnification is defined as the absolute size increase independent of angles.
4. A convex lens of smaller and smaller focal length can infinitely achieve greater magnifying power without physical issues.

Answer Review

1 Consider the statements on simple microscope magnifying lens. Choose the correct statement:
1. The image formed is virtual and closer than infinity if the object is closer than f.
2. It allows the object to be brought closer to the eye than D.
3. Magnification is defined as the absolute size increase independent of angles.
4. A convex lens of smaller and smaller focal length can infinitely achieve greater magnifying power without physical issues.

A 1, 2

B 1, 3, 4

C 2, 3

D 3, 4

💡 Short Explanation

�� Virtual images are formed when the object is within focal length. �� A microscope allows closer viewing. �� Magnification depends on angular size.

✅ Correct Answer Explanation

Statement 1 is correct because when the object is placed within the focal length, the image formed is virtual and generally lies between the near point and infinity. Statement 2 is correct because a microscope enables viewing an object at an effective distance smaller than the normal near point. Statement 3 is incorrect because magnifying power is based on angular magnification, not absolute size increase. Statement 4 is incorrect because practical limitations such as aberrations, fabrication difficulty, and very short working distances prevent unlimited magnification. Therefore Statements 1 and 2 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statements 3 and 4.
�� Option C → Includes incorrect Statement 3.
�� Option D → Statements 3 and 4 are both incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Identify statements that contradict the definition and practical limits of magnification.

Final Logic:

�� Only Statements 1 and 2 are correct.

🧠 Memory Trick

Microscope = Angular Gain, Not Size Gain.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Simple Microscope

2 Identify the incorrect statement regarding virtual magnification in a simple microscope:

A The image subtends the same angle at the eye as the object if the object is placed at the near point.

B The angular magnification relates to how much closer the object can be brought to the eye than 25 cm.

C The virtual image can be comfortably viewed when formed at distances closer than 25 cm without eye strain.

D Magnification relates to the ratio of the angle subtended by the image to that of the object at D.

💡 Short Explanation

�� Comfortable viewing occurs near D or at infinity. �� Images much closer than D strain the eye. �� Angular magnification is angle-based.

✅ Correct Answer Explanation

A virtual image formed significantly closer than the near point requires excessive accommodation by the eye. Such viewing is uncomfortable and causes eye strain. Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Consistent with angular comparison at the near point.
�� Option B → Correct interpretation of magnification.
�� Option D → Correct definition of angular magnification.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall comfortable viewing conditions of the human eye.

Final Logic:

�� Images closer than D are not comfortably viewed.

🧠 Memory Trick

Comfortable Eye = Near Point or Infinity.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Angular Magnification

3 A simple microscope with focal length f = 5 cm forms an image at the near point D = 25 cm. If a defect in the lens requires the near point to shift to D = 30 cm for a specific user, the new magnification would be

A 6

B 7

C 5

D 1

💡 Short Explanation

�� Near-point magnification formula is used. �� M = 1 + D/f �� Replace D with the new near point.

✅ Correct Answer Explanation

Given: f = 5 cm D = 30 cm For image at the near point: M = 1 + D/f M = 1 + 30/5 M = 1 + 6 M = 7 Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Obtained using D/f only.
�� Option C → Incorrect substitution.
�� Option D → Physically incorrect.

🎯 Strategy

Used

Substitution

Application:

�� Substitute the new near-point distance into the formula.

Final Logic:

�� M = 1 + 30/5 = 7.

🧠 Memory Trick

Near Point ⇒ 1 + D/f.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Magnifying Power of Simple Microscope

4 When setting a simple microscope for an infinity view, the magnification m = D/f is slightly less than the near point setting. The primary advantage of this setup is

A higher linear magnification

B lower spherical aberration

C viewing is more comfortable for the relaxed eye

D the object can be placed exactly at 2D

💡 Short Explanation

�� Final image is formed at infinity. �� Eye remains relaxed. �� Long-duration viewing becomes easier.

✅ Correct Answer Explanation

Although the magnification is slightly smaller than the near-point value, image formation at infinity allows the eye muscles to remain relaxed. This minimizes fatigue and provides comfortable viewing. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnification is actually smaller.
�� Option B → Not the primary reason.
�� Option D → Not related to microscope adjustment.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the advantage of image formation at infinity.

Final Logic:

�� Relaxed-eye viewing is the main benefit.

🧠 Memory Trick

Infinity = Relaxed Eye.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Simple Microscope

5 Choose the correct statements about linear magnification (m) of an optical instrument:
1. m = v/u for a lens setup.
2. It is synonymous with the angular size ratio when the object is at D.
3. If v is negative and equal in magnitude to D, the magnification is 1 + D/f.
4. Linear ratio always assumes a real image is formed.

A 1, 2, 3

B 1, 3

C 2, 3, 4

D 1, 4

💡 Short Explanation

�� Linear magnification uses image and object heights. �� Virtual images also have linear magnification. �� Angular and linear magnifications are different.

✅ Correct Answer Explanation

Statement 1 is correct because: m = v/u for a thin lens. Statement 2 is incorrect because angular magnification and linear magnification are distinct concepts. Statement 3 is correct because when the final image is at the near point, the resulting magnifying power becomes 1 + D/f. Statement 4 is incorrect because virtual images can also have linear magnification. Therefore Statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option A → Statement 2 is incorrect.
�� Option C → Statements 2 and 4 are incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

�� Separate angular magnification from linear magnification.

Final Logic:

�� Only Statements 1 and 3 remain correct.

🧠 Memory Trick

Linear = Height Ratio.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Magnification

6 The angular magnification of a simple microscope, when the image is formed at infinity, is derived by the ratio of the angle subtended by the image (θi) to the angle subtended by the object (θo), which simplifies to

A (h/f) / (h/D)

B (h/v) / (h/u)

C (v/D) / (f/h)

D (f/h) / (D/h)

💡 Short Explanation

�� Small-angle approximation is used. �� Angular magnification compares two angles. �� h cancels out.

✅ Correct Answer Explanation

For image at infinity: θi ≈ h/f θo ≈ h/D Therefore: M = θi/θo = (h/f)/(h/D) = D/f Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Not the angular magnification derivation.
�� Option C → Incorrect dimensional relation.
�� Option D → Inverse of the correct ratio.

🎯 Strategy

Used

Formula Recall

Application:

�� Use the standard derivation of angular magnification.

Final Logic:

�� (h/f)/(h/D) simplifies to D/f.

🧠 Memory Trick

Infinity View = D/f.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Angular Magnification

7 Match List I with List II for the compound microscope.

List I	List II
1. Real, inverted, magnified image	a. Formed by objective lens
2. Virtual, enlarged final image	b. Formed by eyepiece
3. fo	c. Focal length of objective (small)
4. fe	d. Focal length of eyepiece (small)

A 1-a, 2-b, 3-c, 4-d

B 1-b, 2-a, 3-d, 4-c

C 1-c, 2-d, 3-a, 4-b

D 1-d, 2-c, 3-b, 4-a

💡 Short Explanation

�� Objective forms the first image. �� Eyepiece forms the final image. �� fo and fe represent focal lengths.

✅ Correct Answer Explanation

1 → a because the objective forms the real inverted magnified image. 2 → b because the eyepiece forms the virtual enlarged final image. 3 → c because fo is the focal length of the objective. 4 → d because fe is the focal length of the eyepiece. Hence: 1-a, 2-b, 3-c, 4-d Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Objective and eyepiece roles are reversed.
�� Option C → Multiple mismatches.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Concept Matching

Application:

�� Match image formation and focal length symbols.

Final Logic:

�� Only Option A gives correct pairings.

🧠 Memory Trick

Objective First, Eyepiece Final.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Compound Microscope

8 A compound microscope achieves a total magnification of 250. If the objective has fo = 1.0 cm and a tube length of 20 cm, the focal length of the eyepiece fe (assuming D = 25 cm and image at infinity) is

A 2.0 cm

B 1.0 cm

C 5.0 cm

D 2.5 cm

💡 Short Explanation

�� Use microscope magnification formula. �� Substitute known values. �� Solve for fe.

✅ Correct Answer Explanation

M = (L/fo) × (D/fe) 250 = (20/1) × (25/fe) 250 = 500/fe fe = 500/250 fe = 2 cm Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives M = 500.
�� Option C → Gives M = 100.
�� Option D → Gives M = 200.

🎯 Strategy

Used

Substitution

Application:

�� Rearrange the microscope magnification formula.

Final Logic:

�� fe = 2 cm.

🧠 Memory Trick

M = (L/fo)(D/fe).

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Compound Microscope Magnification

9 To obtain large magnification in a compound microscope, the objective role dictates that its focal length fo must be

A as large as practically possible

B exactly equal to the tube length

C small, but practically difficult to make much smaller than 1 cm

D greater than the focal length of the eyepiece fe

💡 Short Explanation

�� Objective magnification is inversely proportional to fo. �� Smaller fo gives larger magnification. �� Practical manufacturing limitations exist.

✅ Correct Answer Explanation

The magnification of the objective lens is approximately: mo = L/fo From this relation: • Smaller fo produces larger objective magnification. • Therefore, for a highly magnifying compound microscope, the objective should have a very small focal length. However, making fo extremely small becomes difficult because of: • Lens fabrication limitations • Optical aberrations • Reduced working distance Hence, in practice, fo is kept small but is difficult to make much smaller than about 1 cm. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → A larger fo decreases objective magnification.
�� Option B → There is no requirement that fo equals the tube length.
�� Option D → The objective generally has a smaller focal length than the eyepiece, not a larger one.

🎯 Strategy

Used

Formula Analysis

Application:

�� Use mo = L/fo and determine how fo affects magnification.

Final Logic:

�� Smaller fo gives larger magnification, making Option C correct.

🧠 Memory Trick

Small fo → Large mo.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Objective Lens of Compound Microscope

10 The eyepiece in a compound microscope acts similarly to a _____ microscope, converting the _____ image from the objective into a virtual, enlarged final image.

A Simple, Real

B Telescope, Virtual

C Simple, Erect

D Reflecting, Diminished

💡 Short Explanation

�� The eyepiece behaves like a simple microscope. �� The objective first forms a real image. �� The eyepiece magnifies this image.

✅ Correct Answer Explanation

In a compound microscope: • The objective lens forms a real, inverted, magnified intermediate image. • This intermediate image acts as the object for the eyepiece. • The eyepiece functions exactly like a simple microscope. • It converts the real intermediate image into a virtual, enlarged final image. Thus the blanks are: Simple microscope and Real image. Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → The eyepiece does not act as a telescope, and the intermediate image is real.
�� Option C → The intermediate image formed by the objective is not erect.
�� Option D → The eyepiece is not a reflecting optical system and does not form a diminished image.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the functions of the objective and eyepiece in a compound microscope.

Final Logic:

�� Eyepiece acts as a simple microscope and magnifies the real image formed by the objective.

🧠 Memory Trick

Objective Forms, Eyepiece Magnifies.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Compound Microscope Construction and Working

11 Identify the incorrect statement about the separation and setup in a compound microscope:

A The tube length L is exactly the physical distance between the optical centers of the two lenses.

B The first image is formed near the first focal point of the eyepiece.

C The distance L separates the second focal point of the objective and the first focal point of the eyepiece.

D Multi-component lenses are used to minimize optical aberrations over these separations.

💡 Short Explanation

�� Tube length is defined between focal points. �� It is not the physical lens separation. �� Aberration correction often uses multiple lens elements.

✅ Correct Answer Explanation

Tube length L in a compound microscope is defined as the distance between: • The second focal point of the objective • The first focal point of the eyepiece It is not the exact physical distance between the optical centers of the lenses. Therefore Statement A is incorrect.

❓ Why Other Options Are Incorrect

�� Option B → Correct. Intermediate image is formed near the eyepiece focal plane.
�� Option C → Correct definition of tube length.
�� Option D → Correct. Multi-element lens systems reduce aberrations.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the precise definition of tube length.

Final Logic:

�� Tube length is measured between focal points, not optical centers.

🧠 Memory Trick

Tube Length = F'o to Fe.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Compound Microscope

12 Choose the correct statements regarding the magnifying power of optical instruments:
1. In a compound microscope, total magnification is severely limited to 9.
2. The total magnification formula m = mo × me applies when lenses compound each other's effects.
3. m = (L/fo) × (D/fe) implies m increases as fo and fe decrease.
4. Total power depends heavily on the illumination of the object.

A 2, 3

B 1, 4

C 2, 4

D 1, 2, 3

💡 Short Explanation

�� Microscope magnification can be much greater than 9. �� Total magnification is the product of component magnifications. �� Smaller focal lengths increase magnification.

✅ Correct Answer Explanation

Statement 1 is incorrect because compound microscopes can achieve magnifications far greater than 9. Statement 2 is correct because: m = mo × me Statement 3 is correct because: m = (L/fo) × (D/fe) As fo and fe decrease, magnification increases. Statement 4 is incorrect because illumination affects brightness, not magnifying power directly. Therefore Statements 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 1 and 4 are incorrect.
�� Option C → Includes incorrect Statement 4.
�� Option D → Includes incorrect Statement 1.

🎯 Strategy

Used

Elimination

Application:

�� Identify statements that confuse brightness with magnification.

Final Logic:

�� Only Statements 2 and 3 are correct.

🧠 Memory Trick

Small fo, Small fe → Large M.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Magnifying Power of Compound Microscope

13 A telescope objective has a focal length of 15 m and an eyepiece of 1.0 cm. When viewing the moon (diameter 3.48 × 10⁶ m, orbit radius 3.8 × 10⁸ m), the angular magnification is

A 1500

B 15

C 150

D 1.5

💡 Short Explanation

�� Telescope magnification depends only on focal lengths. �� Convert units first. �� Use M = fo/fe.

✅ Correct Answer Explanation

Given: fo = 15 m = 1500 cm fe = 1 cm For normal adjustment: M = fo/fe M = 1500/1 M = 1500 Therefore Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect division.
�� Option C → Wrong unit conversion.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

Substitution

Application:

�� Convert focal lengths to the same unit and apply the formula.

Final Logic:

�� M = 1500 cm ÷ 1 cm = 1500.

🧠 Memory Trick

Telescope M = fo/fe.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Astronomical Telescope

14 Consider the statements regarding telescope viewing distant objects. Choose the correct statements:
1. Terrestrial telescopes require an extra pair of inverting lenses.
2. The image formed by the objective is at its second focal point.
3. Magnifying power is fe/fo.
4. The final image is usually inverted relative to the original distant object.

A 1, 2, 4

B 2, 3, 4

C 1, 2

D 1, 4

💡 Short Explanation

�� Objective forms image near focal plane. �� Astronomical telescope gives inverted image. �� Magnifying power is fo/fe.

✅ Correct Answer Explanation

Statement 1 is correct because terrestrial telescopes use erecting lenses. Statement 2 is correct because distant objects produce images near the second focal point of the objective. Statement 3 is incorrect because: M = fo/fe not fe/fo. Statement 4 is correct because astronomical telescopes produce inverted images. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Omits correct Statement 4.
�� Option D → Omits correct Statement 2.

🎯 Strategy

Used

Elimination

Application:

�� Verify the telescope magnification formula.

Final Logic:

�� Statement 3 is the only incorrect statement.

🧠 Memory Trick

Telescope = fo/fe.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Telescope

15 In an astronomical telescope, utilizing an objective of exceptionally large diameter primarily serves to

A invert the final image and correct aberrations

B increase light gathering power and improve resolving power

C decrease the overall tube length

D provide a virtual object for the eyepiece

💡 Short Explanation

�� Aperture determines light collection. �� Resolving power improves with diameter. �� Large objectives help observe faint objects.

✅ Correct Answer Explanation

A large objective diameter: • Collects more light • Improves brightness • Increases resolving power • Allows observation of faint celestial objects Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Diameter does not determine image inversion.
�� Option C → Tube length depends mainly on focal lengths.
�� Option D → Not related to objective diameter.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the role of objective aperture.

Final Logic:

�� Larger diameter improves brightness and resolution.

🧠 Memory Trick

Bigger Objective = Brighter + Sharper.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Astronomical Telescope

16 For a Cassegrain reflecting telescope, the effective tube length can be kept relatively short despite a large focal length due to the

A extremely thin primary mirror

B use of a convex secondary mirror to deflect and focus light

C high refractive index of the mirrors

D absence of an eyepiece

💡 Short Explanation

�� Cassegrain uses two mirrors. �� Secondary mirror is convex. �� Light path is folded.

✅ Correct Answer Explanation

A Cassegrain telescope uses: • A concave primary mirror • A convex secondary mirror The convex secondary reflects light back through a hole in the primary mirror, effectively folding the optical path. This provides a large effective focal length in a compact telescope. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not responsible for focal-length increase.
�� Option C → Reflection does not depend on refractive index.
�� Option D → Eyepiece is still present.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the optical design of the Cassegrain telescope.

Final Logic:

�� Folded light path gives large focal length in a short tube.

🧠 Memory Trick

Cassegrain = Folded Optics.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Cassegrain Telescope

17 Resolving power of an astronomical telescope can be limited by the size of the objective. The largest lens objective in use, which exemplifies the practical limits of this approach, is at

A Keck Observatory in Hawaii

B Kavalur Observatory in Tamil Nadu

C Yerkes Observatory in Wisconsin

D Mt. Palomar telescope in California

💡 Short Explanation

�� Very large lenses are difficult to support. �� Refracting telescopes have practical limits. �� Yerkes houses the largest practical refracting telescope.

✅ Correct Answer Explanation

The famous refracting telescope at Yerkes Observatory has an objective diameter of approximately 1.02 m. This represents the practical upper limit for large refracting telescopes because of: • Edge-support problems • Lens weight • Chromatic aberration Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Keck uses reflecting telescopes.
�� Option B → Not the largest refracting objective.
�� Option D → Palomar uses a reflecting telescope.

🎯 Strategy

Used

Fact Recall

Application:

�� Recall famous observatories and telescope types.

Final Logic:

�� Yerkes contains the largest practical refracting telescope.

🧠 Memory Trick

Yerkes = Largest Lens Telescope.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Limitations of Refracting Telescopes

18 Lenses are problematic for massive telescopes because they suffer from _____ aberration and need edge support, whereas mirrors are free from this defect and can be supported by their _____.

A Spherical, Rims

B Chromatic, Entire back surface

C Achromatic, Foci

D Optical, Reflecting faces

💡 Short Explanation

�� Lenses disperse colours. �� Mirrors do not produce chromatic aberration. �� Mirrors can be fully supported from behind.

✅ Correct Answer Explanation

Large objective lenses suffer from chromatic aberration because refractive index depends on wavelength. Large mirrors do not have this defect and can be supported over their entire back surface. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Main defect is chromatic aberration.
�� Option C → Incorrect terminology.
�� Option D → Not a valid optical description.

🎯 Strategy

Used

Concept Recall

Application:

�� Compare lens and mirror objectives.

Final Logic:

�� Mirrors eliminate chromatic aberration and allow full support.

🧠 Memory Trick

Lens = Colour Error, Mirror = No Colour Error.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Reflecting Telescope

19 Choose the correct statements concerning modern reflecting telescopes:
1. They completely lack chromatic aberration.
2. The objective mirror weighs much less than an equivalent lens.
3. The observer cage occasionally obstructs some incoming light.
4. They cannot be built with diameters larger than 1.02 m.

A 1, 2, 3

B 1, 4

C 2, 3, 4

D 1, 2

💡 Short Explanation

�� Mirrors do not suffer chromatic aberration. �� Reflecting telescopes can be very large. �� Secondary support structures may obstruct light.

✅ Correct Answer Explanation

Statement 1 is correct because mirrors do not produce chromatic aberration. Statement 2 is correct because mirrors are lighter than equivalent lenses. Statement 3 is correct because secondary mirrors or observer cages can block a small fraction of incoming light. Statement 4 is incorrect because reflecting telescopes are routinely built much larger than 1.02 m. Therefore Statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits correct Statements 2 and 3 and includes incorrect Statement 4.
�� Option C → Includes incorrect Statement 4.
�� Option D → Omits correct Statement 3.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement confusing reflecting and refracting telescope limitations.

Final Logic:

�� Statement 4 is false; 1, 2 and 3 are correct.

🧠 Memory Trick

Mirrors Grow Bigger Than Lenses.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Reflecting Telescope

20 Match the telescope components (List I) with their specific design uses in reflecting telescopes (List II).

List I	List II
1. Primary mirror	a. Large concave reflector
2. Secondary mirror	b. Allows deflected light to pass through to the observer
3. Hole in primary mirror	c. Convex deflector to focus incident light
4. Eyepiece	d. Magnifies the image formed by the mirrors

A 1-a, 2-c, 3-b, 4-d

B 1-c, 2-a, 3-d, 4-b

C 1-a, 2-b, 3-c, 4-d

D 1-b, 2-d, 3-a, 4-c

💡 Short Explanation

�� Primary mirror collects light. �� Secondary mirror redirects light. �� Eyepiece magnifies the final image.

✅ Correct Answer Explanation

1 → a because the primary mirror is the large concave reflector. 2 → c because the secondary mirror is a convex reflector that redirects and focuses light. 3 → b because the hole in the primary mirror allows reflected light to reach the observer. 4 → d because the eyepiece magnifies the image formed by the mirror system. Therefore: 1-a, 2-c, 3-b, 4-d Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Primary and secondary mirror roles are interchanged.
�� Option C → Incorrect matching of secondary mirror and hole.
�� Option D → Multiple incorrect pairings.

🎯 Strategy

Used

Concept Matching

Application:

�� Match each telescope component with its function.

Final Logic:

�� Only Option A correctly matches all components.

🧠 Memory Trick

Primary Collects → Secondary Redirects → Hole Passes → Eyepiece Magnifies.

📚 Class XII – Physics Part II – Chapter: Ray Optics and Optical Instruments – Topic: Cassegrain Reflecting Telescope

UET UG Physics Booster Test 3 – Optical Instruments

CUET UG Physics Booster Test 3 - Optical Instruments

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