Refraction at Curved Surfaces, Lens, Power of Lens Quiz1

📘 WORKSHEET: Refraction at Curved Surfaces & Lenses


🧠 SECTION A: THEORY / CONCEPTUAL (10 Questions)

Q1. Define refractive index. What does it signify physically?

Q2. State the sign convention used for spherical lenses.

Q3. What is the difference between a real image and a virtual image?

Q4. Define focal length of a lens.

Q5. What is power of a lens? Write its SI unit.

Q6. What happens to the focal length when the refractive index of lens material increases?

Q7. Why does a convex lens converge light rays?

Q8. What is the relation between radius of curvature and focal length for a thin lens?

Q9. Define magnification in case of a lens.

Q10. Can a concave lens form a real image? Justify your answer.


🔢 SECTION B: NUMERICALS (10 Questions)

Use standard sign convention.


Q11. An object is placed 20 cm in front of a convex lens of focal length 10 cm. Find image distance.


Q12. A concave lens has focal length 15 cm. Find its power.


Q13. An object is placed at 30 cm from a convex lens (f = 15 cm). Find magnification.


Q14. Find the focal length of a lens of power +2 D.


Q15. An image is formed at 40 cm by a convex lens when object is at 20 cm. Find focal length.


Q16. A lens has focal length -25 cm. Identify type and find its power.


Q17. Object at infinity forms image at 25 cm. Identify lens and find focal length.


Q18. An object of height 2 cm forms image of height 6 cm. Find magnification.


Q19. A convex lens produces an image at same distance as object. Find object distance in terms of focal length.


Q20. Two lenses of powers +2 D and +3 D are combined. Find net power.


✅ ANSWER KEY

Theory

  1. Ratio of speed of light in vacuum to medium
  2. Cartesian sign convention
  3. Real: formed by actual rays; Virtual: apparent intersection
  4. Distance between optical centre and focus
  5. P=1/fP = 1/f, unit: Dioptre (D)
  6. Focal length decreases
  7. Refraction bends rays toward principal axis
  8. Depends on lens maker formula
  9. m=hi/ho=v/um = h_i/h_o = v/u
  10. No, always virtual

Numericals

  1. +20 cm
  2. -6.67 D
  3. -1
  4. 0.5 m
  5. 13.33 cm
  6. Concave, -4 D
  7. Convex, 25 cm
  8. +3
  9. 2f2f
  10. +5 D

🧮 KEY FORMULA (IMPORTANT)

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
uu
ff
1f=1u+1v, v28, m1\frac{1}{f}=\frac{1}{u}+\frac{1}{v},\ v\approx28,\ m\approx-1
FCconcave mirrorv = 28 | m = -1

Also:

  • P=1/fP = 1/f(f in meters)
  • m=v/um = v/u

✍️ DETAILED SOLUTIONS


Q11

1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u} 110=1v+1(20)\frac{1}{10} = \frac{1}{v} + \frac{1}{(-20)}1v=110+120=320\frac{1}{v} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20} v=20cmv = 20 \, \text{cm}


Q12

P=1f=10.15=6.67DP = \frac{1}{f} = \frac{1}{-0.15} = -6.67 D

Q13

115=1v+1(30)1v=115+130=110\frac{1}{15} = \frac{1}{v} + \frac{1}{(-30)} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{30} = \frac{1}{10}v=10m=1030=1v = 10 \Rightarrow m = \frac{10}{-30} = -1

Q14

f=1P=12=0.5mf = \frac{1}{P} = \frac{1}{2} = 0.5 \, m


Q15

1f=140+1(20)=1240=140f=13.33cm\frac{1}{f} = \frac{1}{40} + \frac{1}{(-20)} = \frac{1 - 2}{40} = -\frac{1}{40} \Rightarrow f = 13.33 \, cm


Q16

P=10.25=4DP = \frac{1}{-0.25} = -4 D

Q17

Object at infinity → image at focus

f=25cmf = 25 \, cm


Q18

m=hiho=62=3m = \frac{h_i}{h_o} = \frac{6}{2} = 3

Q19

For same distance:

u=v=2fu = v = 2f


Q20

Ptotal=P1+P2=2+3=5DP_{total} = P_1 + P_2 = 2 + 3 = 5D

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