1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer: (d) Clay
2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the centre of curvature
(b) At the centre of curvature
(c) At infinity
(d) Between pole of mirror and principle of focus
Answer: (d) Between pole of mirror and principal focus
3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus
Answer: (b) At twice the focal length
4. A spherical mirror and a thin spherical lens have each a focal length of – 15m. The mirror and the lens are likely to be –
(a) Both concave
(b) Both convex
(c) The mirror is concave and the lens is convex
(d) The mirror is convex and the lens is concave
Answer: (a) Both concave
5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be –
(a) Plane
(b) Concave
(c) Convex
(d) Either plane or convex
Answer: (d) Either plane or convex
6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 50cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm
Answer: (c) A convex lens of focal length 5 cm
Certainly! Here are the solutions in the format you requested: each question written exactly as given in the image, followed immediately by its answer. The answers are precise, exam-ready, and suitable for JK Board students.
7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
(a) Name the type of mirror used in the following situations
(b) Headlights of a car
(c) Side/rear-view mirror of a vehicle
(d) Solar furnace
Support your answer with reason.
Answer:
- Range of object distance: 0 to 15 cm (between pole and focus).
- Nature of image: Virtual and erect.
- Size: Larger than the object (magnified).
- Ray diagram: Draw concave mirror; place object between P and F; rays diverge after reflection; extend backward to form virtual, erect, enlarged image behind the mirror.
(a) Headlights of a car: Concave mirror – converges light into a parallel beam.
(b) Side/rear-view mirror: Convex mirror – gives erect, diminished image with wide field of view.
(c) Solar furnace: Concave mirror – converges sunlight to a focus to produce heat.
8. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Answer:
Yes, the lens will still produce a complete image of the object.
Experimental verification: Cover half of a convex lens with black paper. Use it to form an image of a candle on a screen. A full image (though dimmer) is formed.
Explanation: The uncovered half still refracts rays from all points of the object. Each point of the object sends rays through the uncovered portion, forming a complete image. Only the brightness decreases.
9. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Answer:
- Lens formula: 1/f = 1/v – 1/u
f = +10 cm (converging), u = –25 cm (object distance, negative as per sign convention)
1/10 = 1/v – 1/(–25) → 1/10 = 1/v + 1/25 → 1/v = 1/10 – 1/25 = (5–2)/50 = 3/50 → v = 50/3 ≈ 16.67 cm - Position of image: 16.67 cm on the other side of the lens (real image).
- Magnification (m): m = v/u = (50/3) / (–25) = –2/3 ≈ –0.67
- Size of image: = |m| × object height = (2/3) × 5 = 10/3 ≈ 3.33 cm
- Nature: Real, inverted, and diminished (smaller than object).
- Ray diagram: Draw convex lens; place object beyond 2F (25 cm > 20 cm since 2F = 20 cm); image forms between F and 2F on other side, inverted.
10. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Answer:
- Lens formula: 1/f = 1/v – 1/u
f = –15 cm (concave lens), v = –10 cm (image is virtual, on same side as object)
1/(–15) = 1/(–10) – 1/u → –1/15 = –1/10 – 1/u → Multiply by –1: 1/15 = 1/10 + 1/u → 1/u = 1/15 – 1/10 = (2–3)/30 = –1/30 → u = –30 cm - Object distance: 30 cm from lens (on the same side as image).
- Ray diagram: Draw concave lens; object at 30 cm; image at 10 cm on same side, virtual, erect, diminished.
11. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Answer:
- Mirror formula: 1/f = 1/v + 1/u
f = +15 cm (convex mirror), u = –10 cm (object in front)
1/15 = 1/v + 1/(–10) → 1/15 = 1/v – 1/10 → 1/v = 1/15 + 1/10 = (2+3)/30 = 5/30 = 1/6 → v = +6 cm - Position: 6 cm behind the mirror (virtual image).
- Nature: Virtual, erect, and diminished.
- Magnification: m = –v/u = –6/(–10) = 0.6 → smaller than object.
12. The magnification produced by a plane mirror is +1. What does this mean?
Answer:
- Magnification +1 means:
- Image size = Object size (no magnification).
- Positive sign indicates the image is erect (not inverted).
- For a plane mirror, image is virtual, at same distance behind the mirror as object is in front.
13. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Answer:
- Focal length (f): f = R/2 = 30/2 = +15 cm (convex mirror)
- Mirror formula: 1/f = 1/v + 1/u
u = –20 cm, f = +15 cm
1/15 = 1/v + 1/(–20) → 1/15 = 1/v – 1/20 → 1/v = 1/15 + 1/20 = (4+3)/60 = 7/60 → v = 60/7 ≈ 8.57 cm - Position: 8.57 cm behind the mirror (virtual image).
- Nature: Virtual and erect.
- Magnification (m): m = –v/u = –(60/7)/(–20) = (60/7)/20 = 60/(140) = 3/7 ≈ 0.43
- Size of image: = m × object height = (3/7) × 5 = 15/7 ≈ 2.14 cm
- Image smaller than object.
14. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.
Answer:
- Mirror formula: 1/f = 1/v + 1/u
f = –18 cm (concave mirror), u = –27 cm
1/(–18) = 1/v + 1/(–27) → –1/18 = 1/v – 1/27 → 1/v = –1/18 + 1/27 = (–3+2)/54 = –1/54 → v = –54 cm - Screen distance: 54 cm in front of the mirror (real image, so screen placed there).
- Magnification (m): m = –v/u = –(–54)/(–27) = +54/(–27) = –2
- Size of image: = |m| × object height = 2 × 7 = 14 cm
- Nature: Real, inverted, and magnified (larger than object).
15. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer:
- Formula: Power (P) = 1 / f (in meters)
–2.0 = 1/f → f = –1/2 = –0.5 m = –50 cm - Focal length: –50 cm
- Type of lens: Concave lens (negative focal length means diverging lens).
16. A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Answer:
- Formula: P = 1/f (in meters)
+1.5 = 1/f → f = 1/1.5 = 2/3 ≈ 0.667 m = 66.7 cm - Focal length: 66.7 cm (positive)
- Type of lens: Converging lens (convex lens, because positive power corrects hypermetropia/farsightedness).