SET 1

BASIC OPTICS : SHELLS LAW, IMAGE FORMATION BY PLANE MIRROR, CONCAVE MIRROR, CONVEX MIRROR

1. What does Snell’s Law relate?

a) The speed of light in different media
b) The angle of incidence and the angle of refraction
c) The frequency of light waves
d) The wavelength of light in a medium

Answer: b) The angle of incidence and the angle of refraction

Explanation: Snell’s Law is a formula that describes how light bends when passing from one medium to another. It is given by the equation n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2, where n1n_1 and n2n_2 are the refractive indices of the two media, and θ1\theta_1 and θ2\theta_2 are the angles of incidence and refraction.

2. What is the angle of refraction when a light ray strikes a plane surface at an angle of 0°?

a) 30°
b) 45°
c) 0°
d) 90°

Answer: c) 0°

Explanation: When light strikes a surface at an angle of 0° (perpendicular to the surface), there is no bending of light. Thus, the angle of refraction is also 0°.

3. What is the image formed by a plane mirror?

a) Real, inverted
b) Virtual, upright
c) Real, upright
d) Virtual, inverted

Answer: b) Virtual, upright

Explanation: A plane mirror forms a virtual image that is upright and the same size as the object. The image is formed behind the mirror and cannot be projected onto a screen.

4. In Snell’s Law, what happens when light passes from a medium with a higher refractive index to a medium with a lower refractive index?

a) The light bends towards the normal
b) The light bends away from the normal
c) The light stops
d) The light refracts back into the first medium

Answer: b) The light bends away from the normal

Explanation: When light passes from a more optically dense medium (higher refractive index) to a less optically dense medium (lower refractive index), it bends away from the normal.

5. Which of the following is true for a concave mirror?

a) It always forms a real image
b) It forms a virtual image only
c) It can form both real and virtual images
d) It always forms an upright image

Answer: c) It can form both real and virtual images

Explanation: A concave mirror can form both real and virtual images, depending on the position of the object. If the object is placed closer than the focal point, the image will be virtual and upright; otherwise, it will be real and inverted.

6. For a convex mirror, the image formed is always:

a) Real and inverted
b) Virtual and upright
c) Virtual and inverted
d) Real and upright

Answer: b) Virtual and upright

Explanation: A convex mirror always forms a virtual, upright, and diminished image, regardless of the object’s position.

7. What is the focal point of a concave mirror?

a) The point where parallel rays of light converge
b) The point where light rays diverge
c) The point halfway between the mirror and the object
d) The point where the mirror’s axis meets the object

Answer: a) The point where parallel rays of light converge

Explanation: The focal point of a concave mirror is the point where parallel rays of light that are incident on the mirror converge after reflection.

8. A ray of light strikes a convex mirror at an angle of 30°. What is the angle of reflection?

a) 60°
b) 30°
c) 0°
d) 90°

Answer: b) 30°

Explanation: The angle of reflection is always equal to the angle of incidence, regardless of the mirror type. So, if the incident angle is 30°, the reflected angle will also be 30°.

9. What happens when light passes from air (refractive index = 1.0) to water (refractive index = 1.33)?

a) The light speeds up
b) The light bends away from the normal
c) The light bends towards the normal
d) The light is reflected

Answer: c) The light bends towards the normal

Explanation: Since water has a higher refractive index than air, the light slows down and bends towards the normal when entering the water.

10. What is the magnification of an object placed at twice the focal length in front of a concave mirror?

a) 1
b) -1
c) 2
d) -2

Answer: b) -1

Explanation: When an object is placed at twice the focal length of a concave mirror, the image formed is real, inverted, and the same size as the object, leading to a magnification of -1.

11. What type of image does a concave mirror form when the object is placed at the focal point?

a) Virtual and upright
b) Real and inverted
c) Real, inverted, and highly magnified
d) No image is formed

Answer: d) No image is formed

Explanation: When the object is at the focal point of a concave mirror, the reflected rays are parallel and do not meet, meaning no image is formed.

12. A concave mirror has a focal length of 10 cm. What is the image distance when an object is placed 15 cm away from the mirror?

a) 30 cm
b) 20 cm
c) -30 cm
d) -20 cm

Answer: b) 20 cm

Explanation: Using the mirror equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f=10f = 10 cm and do=15d_o = 15 cm, solving for did_i gives di=20d_i = 20 cm.

13. What is the nature of the image formed by a plane mirror?

a) Virtual and reduced in size
b) Virtual and magnified
c) Real and magnified
d) Virtual and the same size

Answer: d) Virtual and the same size

Explanation: A plane mirror forms a virtual image that is the same size as the object and located symmetrically behind the mirror.

14. What is the primary use of concave mirrors?

a) To form virtual images for makeup mirrors
b) To form real images for flashlights and headlights
c) To form virtual images in optical illusions
d) To reflect light for solar panels

Answer: b) To form real images for flashlights and headlights

Explanation: Concave mirrors are used in devices like flashlights and headlights to focus light and form real images.

15. Which of the following does not affect the focal length of a lens?

a) The curvature of the lens
b) The refractive index of the material
c) The color of the light passing through
d) The thickness of the lens

Answer: c) The color of the light passing through

Explanation: The focal length of a lens depends on the curvature of the lens, the refractive index of the material, and the thickness, but not on the color of the light.

16. A convex lens has a focal length of 10 cm. Where would an object need to be placed to form a real image at twice the focal length?

a) 5 cm
b) 10 cm
c) 20 cm
d) 40 cm

Answer: c) 20 cm

Explanation: For a real image to form at twice the focal length, the object must be placed at a distance equal to twice the focal length. Therefore, the object should be placed at 20 cm.

17. The image formed by a convex mirror is always:

a) Real and inverted
b) Virtual and upright
c) Real and upright
d) Virtual and inverted

Answer: b) Virtual and upright

Explanation: A convex mirror always forms a virtual, upright, and diminished image, regardless of the object’s position.

18. What is the relationship between the object distance (do), image distance (di), and focal length (f) for a concave mirror?

a) 1f=1do−1di\frac{1}{f} = \frac{1}{do} - \frac{1}{di}
b) 1f=1do+1di\frac{1}{f} = \frac{1}{do} + \frac{1}{di}
c) f=do×dif = do \times di
d) f=do−dif = do - di

Answer: b) 1f=1do+1di\frac{1}{f} = \frac{1}{do} + \frac{1}{di}

Explanation: This is the mirror equation, which relates the object distance, image distance, and focal length of a mirror

.

19. What is the focal length of a convex mirror?

a) Positive
b) Negative
c) Zero
d) Infinite

Answer: b) Negative

Explanation: The focal length of a convex mirror is considered negative because the focal point is virtual and behind the mirror.

20. When light passes from a denser medium to a rarer medium, what happens to the angle of refraction?

a) It decreases
b) It increases
c) It remains constant
d) It becomes zero

Answer: b) It increases

Explanation: When light passes from a denser medium to a rarer medium, the light speeds up and bends away from the normal, causing the angle of refraction to increase.

SET 2

Converging Lenses (Convex Lenses) and Diverging Lenses (Concave Lenses).

1. What type of image is formed by a converging lens when the object is placed beyond twice the focal length?

a) Virtual and upright
b) Virtual and inverted
c) Real, inverted, and diminished
d) Real, inverted, and magnified

Answer: c) Real, inverted, and diminished

Explanation: When the object is placed beyond twice the focal length (2f) of a converging lens, the image formed is real, inverted, and diminished (smaller than the object).

2. In which situation does a diverging lens form an image?

a) Only when the object is at the focal point
b) Always forms a virtual image
c) Always forms a real image
d) When the object is far from the lens

Answer: b) Always forms a virtual image

Explanation: A diverging lens always forms a virtual, upright, and diminished image, regardless of the position of the object.

3. What is the primary characteristic of the image formed by a converging lens when the object is placed at the focal point?

a) The image is virtual and magnified
b) The image is real and highly magnified
c) No image is formed
d) The image is real and diminished

Answer: c) No image is formed

Explanation: When the object is placed at the focal point of a converging lens, the rays of light after refraction become parallel, meaning no image is formed.

4. What happens to the image formed by a converging lens when the object is placed between the focal point and the lens?

a) The image is real and diminished
b) The image is real and magnified
c) The image is virtual and magnified
d) The image is virtual and diminished

Answer: c) The image is virtual and magnified

Explanation: When an object is placed between the focal point and the lens of a converging lens, the image formed is virtual, upright, and magnified.

5. Which of the following is true about the focal length of a converging lens?

a) It is always negative
b) It is always positive
c) It is zero
d) It can be either positive or negative depending on the lens shape

Answer: b) It is always positive

Explanation: A converging lens (like a convex lens) always has a positive focal length.

6. A diverging lens has a focal length of -10 cm. If an object is placed 15 cm from the lens, what is the image distance?

a) -30 cm
b) 30 cm
c) -6 cm
d) 6 cm

Answer: c) -6 cm

Explanation: Use the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}. Substituting f=−10 cmf = -10 \, \text{cm} and do=15 cmd_o = 15 \, \text{cm}, solving for did_i gives di=−6 cmd_i = -6 \, \text{cm}, indicating that the image is virtual and located on the same side as the object.

7. What type of image does a converging lens always form when the object is placed at infinity?

a) Real and upright
b) Real and diminished
c) Virtual and magnified
d) Real and highly magnified

Answer: b) Real and diminished

Explanation: When the object is at infinity, the image formed by a converging lens will be real, diminished, and located at the focal point.

8. Which of the following is true for a diverging lens?

a) It always forms a virtual, upright, and diminished image
b) It forms a real, inverted, and diminished image
c) It can form both real and virtual images depending on the object position
d) It always forms a real and magnified image

Answer: a) It always forms a virtual, upright, and diminished image

Explanation: A diverging lens always forms a virtual, upright, and diminished image, regardless of the object’s position.

9. What happens to the image formed by a converging lens when the object is at twice the focal length (2f)?

a) The image is virtual and upright
b) The image is real, inverted, and the same size as the object
c) The image is real, inverted, and magnified
d) The image is virtual and magnified

Answer: b) The image is real, inverted, and the same size as the object

Explanation: When the object is placed at twice the focal length (2f) of a converging lens, the image formed is real, inverted, and the same size as the object.

10. A diverging lens has a focal length of -5 cm. If an object is placed 10 cm from the lens, where is the image formed?

a) 10 cm behind the lens
b) 5 cm in front of the lens
c) 20 cm behind the lens
d) The image is not formed

Answer: b) 5 cm in front of the lens

Explanation: Using the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f=−5 cmf = -5 \, \text{cm} and do=10 cmd_o = 10 \, \text{cm}, solving for did_i gives di=−5 cmd_i = -5 \, \text{cm}, which means the image is virtual and located 5 cm in front of the lens.

11. What is the magnification of a diverging lens with an object placed at 30 cm, if the focal length is -15 cm?

a) 0.5
b) -0.5
c) -2
d) 2

Answer: b) -0.5

Explanation: Using the magnification formula M=−didoM = \frac{-d_i}{d_o}, first calculate the image distance using the lens equation. For f=−15 cmf = -15 \, \text{cm} and do=30 cmd_o = 30 \, \text{cm}, solving for did_i gives di=−10 cmd_i = -10 \, \text{cm}. Then, magnification is M=−(−10)30=0.5M = \frac{-(-10)}{30} = 0.5, but the image is virtual and upright, so magnification is -0.5.

12. What is the image distance when the object is at 5 cm and the focal length of a converging lens is 10 cm?

a) 3.33 cm
b) 7.5 cm
c) 15 cm
d) 50 cm

Answer: a) 3.33 cm

Explanation: Using the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f=10 cmf = 10 \, \text{cm} and do=5 cmd_o = 5 \, \text{cm}, solving for did_i gives di=3.33 cmd_i = 3.33 \, \text{cm}. This means the image is real and located at 3.33 cm on the opposite side of the lens.

13. When an object is placed at the focal point of a converging lens, the image formed is:

a) Virtual and magnified
b) Real and diminished
c) Virtual and diminished
d) No image is formed

Answer: d) No image is formed

Explanation: When the object is placed at the focal point of a converging lens, the rays after refraction become parallel, so no image is formed.

14. If the focal length of a diverging lens is -20 cm, where will the image be formed when the object is placed at 40 cm?

a) -40 cm
b) 40 cm
c) -10 cm
d) 10 cm

Answer: c) -10 cm

Explanation: Using the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where f=−20 cmf = -20 \, \text{cm} and do=40 cmd_o = 40 \, \text{cm}, solving for did_i gives di=−10 cmd_i = -10 \, \text{cm}, indicating that the image is virtual and located 10 cm in front of the lens.

15. What happens to the size of the image when an object is moved closer to a converging lens than its focal point?

a) The image becomes real and smaller
b) The image becomes virtual and magnified
c) The image remains the same size
d) The image becomes virtual and diminished

Answer: b) The image becomes virtual and magnified

Explanation: When the object is moved closer to the focal point of a converging lens, the image becomes virtual, upright, and magnified.

16. **What happens to the focal length of a conver

ging lens if the refractive index of the material increases?**

a) It decreases
b) It increases
c) It stays the same
d) It becomes zero

Answer: a) It decreases

Explanation: The focal length of a converging lens decreases as the refractive index of the material increases, according to the lens maker’s equation.

17. Which type of lens is used to correct farsightedness (hyperopia)?

a) Concave lens (diverging)
b) Convex lens (converging)
c) Biconvex lens
d) Plano-convex lens

Answer: b) Convex lens (converging)

Explanation: A convex lens is used to correct farsightedness (hyperopia) because it converges light before it reaches the retina, bringing the image into focus.

18. What is the nature of the image formed by a diverging lens when the object is placed at any distance?

a) Real, inverted, and magnified
b) Virtual, upright, and diminished
c) Real, upright, and magnified
d) Virtual, inverted, and diminished

Answer: b) Virtual, upright, and diminished

Explanation: A diverging lens always forms a virtual, upright, and diminished image, regardless of the object’s position.

19. A converging lens forms an image at a distance of 15 cm when the object is placed at 30 cm. What is the focal length of the lens?

a) 10 cm
b) 5 cm
c) 20 cm
d) 30 cm

Answer: a) 10 cm

Explanation: Using the lens equation 1f=1do+1di\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}, where do=30 cmd_o = 30 \, \text{cm} and di=15 cmd_i = 15 \, \text{cm}, solving for ff gives f=10 cmf = 10 \, \text{cm}.

20. Which of the following describes the image formed by a diverging lens?

a) It is real, inverted, and magnified
b) It is virtual, upright, and diminished
c) It is real, upright, and diminished
d) It is virtual, inverted, and magnified

Answer: b) It is virtual, upright, and diminished

Explanation: A diverging lens always forms a virtual, upright, and diminished image, regardless of the object’s position.

SET 3

Nuclear Radiation

1. Which of the following is a type of nuclear radiation?
   A. Alpha particles
   B. Gamma rays
   C. Beta particles
   D. All of the above
   Answer: D. All of the above
   Explanation: Nuclear radiation includes alpha particles, beta particles, and gamma rays, all of which are emitted during radioactive decay.

2. What is the charge of an alpha particle?
   A. Neutral
   B. Positive (+2)
   C. Negative (-1)
   D. Positive (+1)
   Answer: B. Positive (+2)
   Explanation: An alpha particle is a helium nucleus consisting of two protons and two neutrons, giving it a +2 charge.

3. Beta radiation consists of:
   A. Helium nuclei
   B. Electrons or positrons
   C. Neutrons
   D. Photons
   Answer: B. Electrons or positrons
   Explanation: Beta radiation involves the emission of an electron (β-) or a positron (β+) during radioactive decay.

4. What type of radiation is the most penetrating?
   A. Alpha
   B. Beta
   C. Gamma
   D. Neutron
   Answer: C. Gamma
   Explanation: Gamma rays are high-energy electromagnetic waves that penetrate most materials and require dense shielding like lead to block them.

5. Gamma radiation is best described as:
   A. A stream of particles
   B. High-energy electromagnetic waves
   C. A type of sound wave
   D. A flow of protons
   Answer: B. High-energy electromagnetic waves
   Explanation: Gamma radiation consists of photons, which are high-energy waves with no mass or charge.

6. Which material can effectively block alpha particles?
   A. Paper
   B. Aluminum
   C. Lead
   D. Concrete
   Answer: A. Paper
   Explanation: Alpha particles have low penetrating power and can be stopped by a sheet of paper or even the outer layer of human skin.

7. The half-life of a radioactive isotope is:
   A. The time it takes for the sample to completely decay
   B. The time it takes for half the nuclei in a sample to decay
   C. The time for the isotope to lose its charge
   D. The time it takes to stop emitting radiation
   Answer: B. The time it takes for half the nuclei in a sample to decay
   Explanation: Half-life is the time required for half of the radioactive nuclei in a sample to decay.

8. Beta particles are emitted when:
   A. A neutron turns into a proton and emits an electron
   B. A proton turns into a neutron and emits a positron
   C. Both A and B
   D. None of the above
   Answer: C. Both A and B
   Explanation: β- decay occurs when a neutron turns into a proton, emitting an electron, and β+ decay occurs when a proton becomes a neutron, emitting a positron.

9. What is the SI unit for measuring radioactivity?
   A. Gray (Gy)
   B. Sievert (Sv)
   C. Becquerel (Bq)
   D. Coulomb (C)
   Answer: C. Becquerel (Bq)
   Explanation: The becquerel is the SI unit of radioactivity, measuring disintegrations per second.

10. Which of the following is NOT a source of background radiation?
    A. Cosmic rays
    B. Medical imaging
    C. Radon gas
    D. Visible light
    Answer: D. Visible light
    Explanation: Visible light is not ionizing radiation and does not contribute to background radiation.

11. In nuclear fission, what happens to the nucleus of an atom?
    A. It absorbs energy without splitting
    B. It splits into smaller nuclei, releasing energy
    C. It combines with another nucleus
    D. It becomes completely stable
    Answer: B. It splits into smaller nuclei, releasing energy
    Explanation: Fission occurs when a heavy nucleus splits into lighter nuclei, releasing a large amount of energy.

12. Which type of radiation is often associated with nuclear fission reactions?
    A. Alpha particles
    B. Beta particles
    C. Gamma rays
    D. Visible light
    Answer: C. Gamma rays
    Explanation: Gamma radiation is often emitted during nuclear fission, accompanying the release of neutrons.

13. What device is used to measure radiation?
    A. Thermometer
    B. Geiger-Müller counter
    C. Voltmeter
    D. Barometer
    Answer: B. Geiger-Müller counter
    Explanation: A Geiger-Müller counter detects ionizing radiation and measures its intensity.

14. What occurs during nuclear fusion?
    A. Two light nuclei combine to form a heavier nucleus
    B. A heavy nucleus splits into lighter nuclei
    C. Electrons are emitted from the nucleus
    D. A neutron is absorbed by the nucleus
    Answer: A. Two light nuclei combine to form a heavier nucleus
    Explanation: Fusion powers stars by combining hydrogen nuclei into helium, releasing vast energy.

15. What shields gamma radiation most effectively?
    A. Plastic
    B. Aluminum foil
    C. Thick lead
    D. Water
    Answer: C. Thick lead
    Explanation: Gamma rays are best blocked by dense materials like lead or thick concrete.

16. Which particle has the greatest mass?
    A. Alpha particle
    B. Beta particle
    C. Gamma ray
    D. Neutron
    Answer: A. Alpha particle
    Explanation: Alpha particles are helium nuclei, making them the heaviest among the options.

17. What is the role of a moderator in a nuclear reactor?
    A. To increase the speed of neutrons
    B. To slow down neutrons for fission
    C. To absorb excess gamma radiation
    D. To reduce heat generation
    Answer: B. To slow down neutrons for fission
    Explanation: Moderators, such as water or graphite, reduce neutron speeds to sustain the chain reaction.

18. Why is nuclear radiation dangerous to living tissues?
    A. It produces heat inside cells
    B. It ionizes atoms, causing cellular damage
    C. It depletes oxygen in tissues
    D. It creates electrical signals in the body
    Answer: B. It ionizes atoms, causing cellular damage
    Explanation: Radiation can break chemical bonds in cells, leading to mutations or cell death.

19. The Chernobyl disaster was caused by:
    A. A meteor impact
    B. A chemical explosion
    C. A nuclear reactor meltdown
    D. A gamma-ray burst
    Answer: C. A nuclear reactor meltdown
    Explanation: The Chernobyl disaster was a catastrophic failure of a nuclear reactor due to design flaws and operator errors.

20. Which of the following is a common use of radioactive isotopes?
    A. Generating electricity
    B. Medical imaging and treatment
    C. Food sterilization
    D. All of the above
    Answer: D. All of the above
    Explanation: Radioisotopes have multiple applications, including power generation, medical uses, and food preservation.

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