UET UG Physics Booster Test 2 - Optical Instruments -

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CUET UG Physics Booster Test 2 - Optical Instruments

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Physics Booster Test 2 - Optical Instruments

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QUESTION 1 OF 20

To act as a simple microscope, an object should be held relative to the converging lens at a distance that is

Answer Review

1 To act as a simple microscope, an object should be held relative to the converging lens at a distance that is

A equal to twice the focal length

B one focal length away or less

C greater than two focal lengths

D precisely infinity

💡 Short Explanation

�� A simple microscope uses a convex lens. �� The object must be within the focal length. �� A virtual magnified image is produced.

✅ Correct Answer Explanation

For a convex lens to act as a simple microscope, the object must be placed between the optical center and the focal point. That means the object distance must be less than or equal to the focal length. Under this condition, the image formed is: • Virtual • Erect • Magnified Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Produces a real image.
�� Option C → Object beyond 2F does not give magnified virtual image.
�� Option D → Impossible object position for microscope operation.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the object position required for a simple microscope.

Final Logic:

�� Object must be within the focal length to obtain a virtual magnified image.

🧠 Memory Trick

Magnifier = Object Inside Focus.

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

2 The virtual image produced by a simple microscope is most suitable for viewing when it is formed at

A the optical center

B a distance closer than the near point

C infinity for a relaxed eye

D the focal point of the lens

💡 Short Explanation

�� Relaxed eye viewing is preferred. �� Eye strain is minimum. �� Final image is placed at infinity.

✅ Correct Answer Explanation

The most comfortable viewing occurs when the final image is formed at infinity. In this case: • Eye muscles remain relaxed. • Continuous observation is easier. • Angular magnification is D/f. Therefore Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Optical center is not an image position used for viewing.
�� Option B → Closer than near point cannot be viewed comfortably.
�� Option D → Image at focal point is not the final viewing condition.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the condition for comfortable viewing.

Final Logic:

�� Relaxed eye corresponds to image at infinity.

🧠 Memory Trick

Relaxed Eye = Infinity.

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

3 For a simple microscope with a lens power of +10 D, assuming the near point is 25 cm, the maximum magnification achieved by this lens is

A 2.5

B 3.5

C 1.5

D 4.5

💡 Short Explanation

�� Maximum magnification occurs at the near point. �� M = 1 + D/f �� Convert power into focal length first.

✅ Correct Answer Explanation

Given: P = +10 D f = 1/P f = 1/10 m f = 0.1 m = 10 cm Maximum magnification: M = 1 + D/f M = 1 + 25/10 M = 1 + 2.5 M = 3.5 Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Ignores the extra 1 in the formula.
�� Option C → Incorrect calculation.
�� Option D → Overestimation.

🎯 Strategy

Used

Substitution

Application:

�� Convert power to focal length and apply magnification formula.

Final Logic:

�� M = 1 + 25/10 = 3.5.

🧠 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 the image in a simple microscope is formed at infinity, the magnification is obtained as m = D/f. This magnification is

A equal to the magnification at the near point

B one less than the magnification when the image is at the near point

C twice the magnification at the near point

D one greater than the magnification at the near point

💡 Short Explanation

�� Two standard magnification formulas exist. �� Near point gives larger magnification. �� Difference equals 1.

✅ Correct Answer Explanation

For image at infinity: M∞ = D/f For image at near point: MN = 1 + D/f Therefore: MN = M∞ + 1 Hence magnification at infinity is one less than the magnification at the near point. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnifications are not equal.
�� Option C → No factor of two exists.
�� Option D → Reversed relationship.

🎯 Strategy

Used

Substitution

Application:

�� Compare the two magnification formulas.

Final Logic:

�� D/f is exactly one less than 1 + D/f.

🧠 Memory Trick

Infinity Formula Missing +1.

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

5 Identify the incorrect statement regarding the linear magnification of optical instruments.

A It is defined as the ratio of image height to object height.

B It is given by v/u for a single lens.

C For an erect and virtual image formed by a lens, linear magnification is positive.

D The linear magnification is always equal to the angular magnification in telescopes.

💡 Short Explanation

�� Linear and angular magnifications are different concepts. �� Linear magnification deals with heights. �� Telescopes primarily use angular magnification.

✅ Correct Answer Explanation

Statement D is incorrect because telescopes are characterized by angular magnification, not linear magnification. Linear magnification depends on image and object heights. Angular magnification depends on the angles subtended at the eye. The two are generally not equal. Hence Option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct definition of linear magnification.
�� Option B → m = v/u is valid for a lens.
�� Option C → Virtual erect images have positive magnification.

🎯 Strategy

Used

Concept Recall

Application:

�� Distinguish between linear and angular magnification.

Final Logic:

�� Telescopes use angular magnification, making Option D incorrect.

🧠 Memory Trick

Height → Linear, Angle → Angular.

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

6 Choose the correct statements about angular magnification:
1. It allows the object to be brought closer to the eye than D.
2. It is the ratio of the angle subtended by the image to the angle subtended by the object if placed at D.
3. It describes the actual absolute size increase of the object.
4. Without a microscope, a small object at D subtends a very small angle.

A 1, 2, 4

B 2, 3

C 1, 3, 4

D 2, 4

💡 Short Explanation

�� Angular magnification is based on viewing angle. �� It does not represent actual size increase. �� Small objects subtend small angles at D.

✅ Correct Answer Explanation

Statement 1 is correct because optical instruments allow effective viewing at distances smaller than D. Statement 2 is correct because angular magnification is defined using angular comparison. Statement 3 is incorrect because angular magnification does not represent actual physical enlargement. Statement 4 is correct because a small object at D subtends a very small angle. Therefore Statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes incorrect Statement 3.
�� Option C → Includes incorrect Statement 3.
�� Option D → Omits correct Statement 1.

🎯 Strategy

Used

Elimination

Application:

�� Identify the statement confusing angular magnification with physical enlargement.

Final Logic:

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

🧠 Memory Trick

Angular ≠ Actual Size.

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

7 A compound microscope overcomes the limited maximum magnification of a _____ microscope by using a second lens called the _____.

A Simple, Objective

B Simple, Eyepiece

C Refracting, Eyepiece

D Telescope, Objective

💡 Short Explanation

�� Compound microscope improves simple microscope magnification. �� A second lens is added. �� That second lens is the eyepiece.

✅ Correct Answer Explanation

A simple microscope has limited magnification. A compound microscope uses: • Objective lens • Eyepiece lens The eyepiece magnifies the image produced by the objective, increasing total magnification. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Objective is not the second lens added for magnification.
�� Option C → Refracting is not the relevant instrument.
�� Option D → Telescope is unrelated.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the purpose of a compound microscope.

Final Logic:

�� Compound microscope extends simple microscope capability using an eyepiece.

🧠 Memory Trick

Compound = Objective + Eyepiece.

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

8 In a compound microscope, if the objective magnification is 10 and the eyepiece magnification is 5, the total magnification is

A 15

B 50

C 2

D 500

💡 Short Explanation

�� Total magnification is a product. �� Multiply objective and eyepiece magnifications. �� Not an addition.

✅ Correct Answer Explanation

For a compound microscope: M = mo × me Given: mo = 10 me = 5 M = 10 × 5 M = 50 Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Magnifications are not added.
�� Option C → Incorrect calculation.
�� Option D → Excessive multiplication.

🎯 Strategy

Used

Substitution

Application:

�� Apply the magnification formula directly.

Final Logic:

�� 10 × 5 = 50.

🧠 Memory Trick

Microscope Magnification = Product.

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

9 Consider the statements regarding the objective lens in a compound microscope setup. Choose the correct statements:
1. It forms a virtual image.
2. Its focal length should be small to achieve large magnification.
3. The first image is formed near the focal point of the eyepiece.
4. It makes the final image erect.

A 1, 4

B 2, 3

C 1, 2, 3

D 3, 4

💡 Short Explanation

�� Objective forms a real image. �� Objective has small focal length. �� Intermediate image lies near eyepiece focus.

✅ Correct Answer Explanation

Statement 1 is incorrect because the objective forms a real, inverted image. Statement 2 is correct because small focal length gives high magnification. Statement 3 is correct because the intermediate image is formed near the focal plane of the eyepiece. Statement 4 is incorrect because the final image is inverted relative to the object. Therefore Statements 2 and 3 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Identify image characteristics of the objective.

Final Logic:

�� Only Statements 2 and 3 are correct.

🧠 Memory Trick

Objective = Real First Image.

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

10 In a compound microscope, the first inverted image formed by the objective is placed

A at infinity relative to the eyepiece

B near or within the focal plane of the eyepiece

C at the optical center of the eyepiece

D beyond twice the focal length of the eyepiece

💡 Short Explanation

�� Objective forms an intermediate image. �� Eyepiece acts as a magnifier. �� Image is placed near eyepiece focus.

✅ Correct Answer Explanation

The objective forms a real inverted image. This intermediate image is positioned near the first focal point of the eyepiece so that the eyepiece can magnify it effectively. Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Intermediate image is not at infinity.
�� Option C → Optical center is not the required location.
�� Option D → Not used in microscope design.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall image formation in a compound microscope.

Final Logic:

�� Intermediate image must lie near the eyepiece focal plane.

🧠 Memory Trick

Objective Image Near Eyepiece Focus.

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

11 Identify the incorrect statement about the tube length of a compound microscope:

A It is the distance between the second focal point of the objective and the first focal point of the eyepiece.

B A larger tube length L generally decreases the total magnification.

C Large lenses are required to make L large.

D It appears in the numerator of the objective magnification formula mo = L/fo.

💡 Short Explanation

�� Objective magnification depends on tube length. �� Larger L increases objective magnification. �� Total magnification increases with L.

✅ Correct Answer Explanation

For a compound microscope: mo = L/fo Total magnification: M = (L/fo) × (D/fe) Since L appears in the numerator, increasing tube length increases both objective magnification and total magnification. Therefore Statement B is incorrect. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct definition of tube length.
�� Option C → Larger tube lengths generally require larger instrument dimensions.
�� Option D → Correct formula relation.

🎯 Strategy

Used

Formula Analysis

Application:

�� Examine how L appears in the magnification formula.

Final Logic:

�� Larger L increases magnification, not decreases it.

🧠 Memory Trick

Larger L = Larger M.

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

12 Match the parameters of a compound microscope (List I) with their descriptions (List II).

List I	List II
1. fo	a. Tube length
2. L	b. Near point distance
3. fe	c. Focal length of objective
4. D	d. Focal length of eyepiece

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

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

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

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

💡 Short Explanation

�� fo refers to objective focal length. �� fe refers to eyepiece focal length. �� D is near point distance.

✅ Correct Answer Explanation

1 → c because fo is focal length of objective. 2 → a because L represents tube length. 3 → d because fe is focal length of eyepiece. 4 → b because D denotes near point distance. Therefore: 1-c, 2-a, 3-d, 4-b Hence Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect assignment of fo and L.
�� Option C → Multiple incorrect matches.
�� Option D → Incorrect matching of L and fe.

🎯 Strategy

Used

Concept Matching

Application:

�� Match each symbol with its standard meaning.

Final Logic:

�� Only Option A correctly matches all parameters.

🧠 Memory Trick

fo = Objective, fe = Eyepiece.

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

13 An astronomical telescope has an objective of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is its magnifying power in normal adjustment?

A 150

B 24

C 138

D 864

💡 Short Explanation

�� Telescope magnification depends on focal lengths. �� M = fo/fe �� Use normal adjustment formula.

✅ Correct Answer Explanation

For a telescope in normal adjustment: M = fo/fe Given: fo = 144 cm fe = 6 cm M = 144/6 M = 24 Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect division.
�� Option C → Subtraction used incorrectly.
�� Option D → Multiplication used incorrectly.

🎯 Strategy

Used

Substitution

Application:

�� Apply the telescope magnification formula directly.

Final Logic:

�� M = 144/6 = 24.

🧠 Memory Trick

Telescope M = fo/fe.

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

14 Choose the correct statements about viewing distant objects with a telescope:
1. Light from a distant object forms a real image at the objective's second focal point.
2. The final image is erect relative to the original object.
3. Terrestrial telescopes use an extra pair of inverting lenses.
4. The objective must have a very small focal length.

A 1, 3

B 2, 4

C 1, 2, 3

D 3, 4

💡 Short Explanation

�� Objective forms a real image. �� Terrestrial telescopes use erecting lenses. �� Telescope objective has large focal length.

✅ Correct Answer Explanation

Statement 1 is correct because parallel rays from distant objects form a real image at the objective focal plane. Statement 2 is incorrect because an astronomical telescope produces an inverted final image. Statement 3 is correct because terrestrial telescopes use additional erecting lenses. Statement 4 is incorrect because telescope objectives require large focal lengths. Therefore Statements 1 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Statements 2 and 4 are incorrect.
�� Option C → Includes incorrect Statement 2.
�� Option D → Includes incorrect Statement 4.

🎯 Strategy

Used

Elimination

Application:

�� Identify the incorrect statements regarding telescope image orientation and focal length.

Final Logic:

�� Only Statements 1 and 3 are correct.

🧠 Memory Trick

Astronomy: Inverted, Terrestrial: Erected.

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

15 Telescope large objective statements are provided below. Choose the correct statements:
1. It gathers more light to observe fainter objects.
2. It decreases the resolving power of the telescope.
3. It increases the resolving power of the telescope.
4. It creates a wider aperture.

A 1, 2, 4

B 1, 3, 4

C 1, 4

D 2, 3

💡 Short Explanation

�� Larger objective collects more light. �� Resolving power improves. �� Aperture increases.

✅ Correct Answer Explanation

Statement 1 is correct because a larger objective gathers more light. Statement 2 is incorrect because larger apertures increase, not decrease, resolving power. Statement 3 is correct because resolving power improves with objective diameter. Statement 4 is correct because larger objectives provide larger apertures. Therefore Statements 1, 3 and 4 are correct.

❓ Why Other Options Are Incorrect

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

🎯 Strategy

Used

Elimination

Application:

�� Evaluate the effect of aperture on resolving power.

Final Logic:

�� Large objective improves both light gathering and resolution.

🧠 Memory Trick

Bigger Aperture = Brighter + Sharper.

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

16 For a telescope in normal adjustment, if the objective and eyepiece focal lengths are 100 cm and 5 cm respectively, the tube length is

A 20 cm

B 105 cm

C 95 cm

D 500 cm

💡 Short Explanation

�� Normal adjustment means image at infinity. �� Tube length equals sum of focal lengths. �� Add fo and fe.

✅ Correct Answer Explanation

For normal adjustment: Length = fo + fe Given: fo = 100 cm fe = 5 cm Length = 100 + 5 Length = 105 cm Therefore Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Division performed incorrectly.
�� Option C → Difference used incorrectly.
�� Option D → Product used incorrectly.

🎯 Strategy

Used

Substitution

Application:

�� Apply the normal-adjustment length formula.

Final Logic:

�� 100 + 5 = 105 cm.

🧠 Memory Trick

Telescope Length = fo + fe.

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

17 The desirable aim in optical astronomical telescopes to improve their resolution is to

A use objectives of very small diameter

B make them with objectives of large diameter

C use an eyepiece with a large focal length

D decrease the tube length significantly

💡 Short Explanation

�� Resolution depends on aperture. �� Larger aperture reduces diffraction. �� Objective diameter is crucial.

✅ Correct Answer Explanation

Resolving power increases with objective diameter. A larger aperture: • Collects more light. • Reduces diffraction effects. • Improves separation of nearby objects. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Small diameter worsens resolution.
�� Option C → Eyepiece focal length does not determine resolving power.
�� Option D → Tube length does not significantly determine resolution.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall factors affecting resolving power.

Final Logic:

�� Large objective diameter improves resolution.

🧠 Memory Trick

Large Aperture = High Resolution.

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

18 Big lenses in refracting telescopes are difficult to support by their _____ and form images that suffer from _____ aberration.

A Centers, Spherical

B Edges, Chromatic

C Focal points, Astigmatic

D Poles, Optic

💡 Short Explanation

�� Large lenses are supported at edges. �� Chromatic aberration is a major problem. �� Reflecting telescopes overcome this issue.

✅ Correct Answer Explanation

Large objective lenses are generally supported only at their edges. This creates mechanical difficulties. In addition, lenses disperse light and therefore suffer from chromatic aberration. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Support is not through the center.
�� Option C → Not the standard telescope limitation.
�� Option D → Not a valid optical description.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall limitations of refracting telescopes.

Final Logic:

�� Edge support and chromatic aberration are key issues.

🧠 Memory Trick

Lens = Edge Support + Colour Error.

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

19 A major mechanical advantage of using a mirror objective over a lens objective in large telescopes is that the mirror

A cannot be scratched

B can be supported over its entire back surface

C is strictly transparent

D requires complex inversion lenses

💡 Short Explanation

�� Mirrors are mechanically easier to support. �� Large mirrors do not sag significantly. �� This allows construction of larger telescopes.

✅ Correct Answer Explanation

Large mirrors can be supported over their entire rear surface. This provides: • Better stability • Less deformation • Easier construction of large telescopes Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Mirrors can still be scratched.
�� Option C → Mirrors are reflective, not transparent.
�� Option D → Not an advantage of mirrors.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the structural advantages of reflecting telescopes.

Final Logic:

�� Full back support is the major mechanical advantage.

🧠 Memory Trick

Mirror Supported from Behind.

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

20 Identify the incorrect statement about the Cassegrain telescope:

A It deflects light being focused using a secondary mirror.

B The secondary mirror is concave in shape.

C The light passes through a hole in the objective primary mirror.

D It provides a large focal length in a short telescope.

💡 Short Explanation

�� Cassegrain telescope uses two mirrors. �� Secondary mirror is convex. �� Compact design gives long focal length.

✅ Correct Answer Explanation

A Cassegrain telescope contains: • A large concave primary mirror. • A small convex secondary mirror. The secondary mirror reflects light back through a central hole in the primary mirror. Therefore Statement B is incorrect. Hence Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct description.
�� Option C → Correct feature of Cassegrain design.
�� Option D → Major advantage of the design.

🎯 Strategy

Used

Concept Recall

Application:

�� Recall the structure of a Cassegrain telescope.

Final Logic:

�� The secondary mirror is convex, not concave.

🧠 Memory Trick

Cassegrain = Concave Primary + Convex Secondary.

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

UET UG Physics Booster Test 2 – Optical Instruments

CUET UG Physics Booster Test 2 - Optical Instruments

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