Light - Reflection and Refraction — Class 10 Science practice questions

Question 1

At the Jawaharlal Nehru Planetarium in Bengaluru, Karnataka, a technician named Arvind is inspecting a large concave mirror used in the projection system. The mirror has a radius of curvature of 40 cm. During a routine check, Arvind places a small luminous object at a distance of 30 cm in front of the mirror along its principal axis to verify the optical properties of the mirror before the evening show. The size of the object placed is 2 cm.

Accepted Answer

**(i)**

Since f = R/2,

f = 40/2 = 20 cm

Applying the sign convention for a concave mirror, **f = −20 cm**.

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**(ii)**

Given: u = −30 cm, f = −20 cm

Using the mirror formula:

1/v + 1/u = 1/f

⟹ 1/v + 1/(−30) = 1/(−20)

⟹ 1/v = −1/20 + 1/30 = (−3 + 2)/60 = −1/60

∴ **v = −60 cm**

The negative sign indicates that the image is formed **60 cm in front of the mirror** on the same side as the object. The image is **real**.

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**(iii)**

Given: v = −60 cm, u = −30 cm, h₀ = 2 cm

Magnification: m = −v/u = −(−60)/(−30) = −2

Size of image: hᵢ = m × h₀ = −2 × 2 = **−4 cm**

The magnitude of the image size is **4 cm**. The negative sign confirms the image is **inverted**. Since |m| = 2 > 1, the image is **magnified**.

∴ The image formed is **real, inverted, and magnified**, with a size of 4 cm, located 60 cm in front of the mirror.

Question 2

A science club at a school set up a demonstration using a concave mirror of radius of curvature 30 cm. They placed a bright LED bulb at different positions along the principal axis and recorded the nature of images formed on a screen kept on the same side as the reflected rays. During one observation, the LED was placed 10 cm in front of the mirror. A junior student, looking at the screen, found no image on it and was confused. The teacher then asked the junior student to remove the screen and look into the mirror directly.

(a) What is the focal length of the concave mirror used in the demonstration?

(b) When the LED is placed 10 cm in front of the mirror, where does the image form, and why can it NOT be obtained on a screen?

(c) Calculate the position and magnification of the image when the LED is placed 45 cm in front of the same mirror. State the nature of the image formed.

OR

(c) The club then replaced the concave mirror with a convex mirror of the same radius of curvature (30 cm) and placed the LED 45 cm in front of it. Calculate the position of the image and state TWO characteristics of the image formed.

Accepted Answer

(a) Focal length of the concave mirror:

Using the relation f = R/2,

f = 30/2 = 15 cm

Since the mirror is concave, f = −15 cm (using the New Cartesian Sign Convention).

[1 mark]

(b) Position of the image and reason it cannot be obtained on a screen:

The LED is placed 10 cm in front of the mirror, i.e., between the pole (P) and the focus (F), since f = 15 cm and the object distance is only 10 cm.

For a concave mirror, when the object is placed between P and F, the reflected rays diverge after reflection. These diverging rays, when extended behind the mirror, appear to meet at a point behind the mirror.

Therefore, the image is formed BEHIND the mirror (on the non-reflecting side).

The image cannot be obtained on a screen because it is a VIRTUAL image — the reflected rays do not actually meet at a point; they only appear to diverge from a point behind the mirror. A screen can only receive real images where rays actually converge.

[1 mark]

(c) Calculation when LED is placed 45 cm in front of the concave mirror:

Given:
Using sign convention: u = −45 cm, f = −15 cm

Using the mirror formula:
1/v + 1/u = 1/f

⟹ 1/v = 1/f − 1/u
⟹ 1/v = 1/(−15) − 1/(−45)
⟹ 1/v = −1/15 + 1/45
⟹ 1/v = (−3 + 1)/45
⟹ 1/v = −2/45
⟹ v = −22.5 cm

∴ The image is formed 22.5 cm in front of the mirror (on the same side as the object).

Magnification:
m = −v/u = −(−22.5)/(−45) = −22.5/45 = −0.5

Nature of the image:
- v is negative → image is in front of the mirror → REAL
- m is negative → image is INVERTED
- |m| = 0.5 < 1 → image is DIMINISHED

∴ The image is real, inverted and diminished.

[2 marks]

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OR
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(c) Calculation when the LED is placed 45 cm in front of a CONVEX mirror (R = 30 cm):

For a convex mirror:
f = +R/2 = +30/2 = +15 cm
Using sign convention: u = −45 cm, f = +15 cm

Using the mirror formula:
1/v + 1/u = 1/f

⟹ 1/v = 1/f − 1/u
⟹ 1/v = 1/15 − 1/(−45)
⟹ 1/v = 1/15 + 1/45
⟹ 1/v = (3 + 1)/45
⟹ 1/v = 4/45
⟹ v = +11.25 cm

∴ The image is formed 11.25 cm behind the mirror.

TWO characteristics of the image:
(i) The image is VIRTUAL and ERECT — v is positive, meaning the image forms behind the mirror where reflected rays cannot actually meet; the image can only be seen by looking into the mirror.
(ii) The image is DIMINISHED — for a convex mirror, the image is always smaller than the object regardless of object position.

[2 marks]

Question 3

Riya is decorating her room and places a small decorative candle 30 cm in front of a concave mirror. She notices that the mirror produces a clear, sharp image of the candle flame on the wall behind her, exactly 90 cm from the mirror. She measures the height of the candle as 2 cm.

(a) What type of image is formed on the wall — real or virtual? Give one reason for your answer.

(b) Calculate the focal length of the concave mirror used.

(c) Calculate the size of the image of the candle formed on the wall. Also state whether the image is magnified or diminished.

Accepted Answer

(a) The image formed on the wall is a REAL image.
Reason: The image is formed on a screen (the wall), which is on the same side as the object in front of the mirror. Only a real image can be obtained on a screen; a virtual image cannot be projected onto a screen.
[1 mark]

(b) Using the New Cartesian Sign Convention:
Object distance, u = −30 cm (object in front of mirror, against incident light)
Image distance, v = −90 cm (image is in front of mirror, on same side as object — real image)

Using the mirror formula:
1/f = 1/v + 1/u
1/f = 1/(−90) + 1/(−30)
1/f = −1/90 − 1/30
1/f = −1/90 − 3/90
1/f = −4/90
1/f = −2/45
∴ f = −45/2 = −22.5 cm

The focal length of the concave mirror is 22.5 cm.
[1 mark]

(c) Using the magnification formula for a mirror:
m = −v/u
m = −(−90)/(−30)
m = −90/30
m = −3

Size of image = m × size of object
Height of image = |m| × height of candle
Height of image = 3 × 2 = 6 cm

Since |m| = 3 > 1, the image is MAGNIFIED.
Since m is negative, the image is real and inverted.
∴ The image of the candle is 6 cm tall, real, inverted, and magnified.
[2 marks]

Question 4

Rohan is sitting in a swimming pool and looks up at a coin placed on the edge of the pool deck. His friend Meera, standing outside the pool, looks at a fish swimming below the water surface. A lifeguard at the pool has placed a convex lens (focal length 20 cm) near the pool entrance to use as a magnifying glass to read fine-print safety notices.

Based on this situation, answer the following:
(a) Meera looks at the fish which is actually 60 cm below the water surface. If the refractive index of water is 4/3, what is the apparent depth at which she sees the fish? [1 mark]
(b) Rohan notices that the coin on the pool deck appears to be at a different position than it actually is. State the phenomenon responsible for this observation and in which direction (closer or farther) does the coin appear to him. [1 mark]
(c) The lifeguard places the safety notice 15 cm in front of the convex lens (focal length 20 cm). Using the lens formula, find the position and nature of the image formed. [2 marks]

Accepted Answer

(a) Apparent depth = Real depth / refractive index

Apparent depth = 60 / (4/3) = 60 × (3/4) = 45 cm

∴ Meera sees the fish at an apparent depth of 45 cm below the surface. [1 mark]

(b) The phenomenon responsible is Refraction of light. When light travels from a denser medium (water) to a rarer medium (air), it bends away from the normal. As a result, Rohan sees the coin appearing to be farther away (at a greater distance) than its actual position, because the refracted rays appear to diverge from a point higher up than the actual object. [1 mark]

(c) Given:
Object distance, u = −15 cm (using sign convention; object is on the left of lens)
Focal length of convex lens, f = +20 cm

Using the lens formula:
1/v − 1/u = 1/f

1/v = 1/f + 1/u
1/v = 1/20 + 1/(−15)
1/v = 1/20 − 1/15
1/v = (3 − 4)/60
1/v = −1/60

∴ v = −60 cm

The negative sign indicates that the image is formed on the same side as the object (i.e., to the left of the lens).

Magnification: m = v/u = (−60)/(−15) = +4

Positive magnification confirms the image is erect.

∴ The image is formed 60 cm on the same side as the object. It is virtual, erect, and magnified (4 times the size of the notice). [1 + 1 mark]

Question 5

Riya is setting up a science project at home. She places a shiny steel spoon in front of her face and notices that when she looks at the concave side of the spoon, her face appears large and upright, but when she turns the spoon and looks at the convex side, her face appears small and upright. Curious, she then holds the concave side close to a wall on a sunny day and notices a bright spot of light on the wall.

Based on the above situation, answer the following questions:
(a) Name the type of mirror that forms a virtual, erect and magnified image of an object placed close to it. [1]
(b) When Riya looks at the convex side of the spoon, the image of her face is always virtual, erect and diminished regardless of her distance from the spoon. Give one reason why this is so. [1]
(c) When Riya holds the concave side of the spoon towards sunlight and a bright spot forms on the wall, what is that bright spot called, and at what specific point does it form? Also explain why parallel rays from the Sun converge to this point after reflection. [2]

Accepted Answer

(a) Concave mirror forms a virtual, erect and magnified image when the object is placed between the pole (P) and the principal focus (F) of the mirror.

(b) A convex mirror always diverges the reflected rays. The reflected rays, when extended behind the mirror, appear to meet at a point behind the mirror. Since the reflected rays never actually meet in front of the mirror, the image is always virtual and erect. The diverging nature of a convex mirror means the image is always formed between the pole and the principal focus (behind the mirror), and is always smaller than the object — regardless of the object's distance from the mirror.

(c) The bright spot formed on the wall is called the Principal Focus (or Focal Point) of the concave mirror. It forms at a point in front of the mirror at a distance equal to the focal length (f) from the pole.

The Sun is approximately 150 million km away, so rays coming from it are effectively parallel to the principal axis when they reach the spoon. According to the property of a concave mirror, all rays travelling parallel to the principal axis, after reflection, converge at the principal focus. This is because the concave mirror is a part of a hollow sphere; each point on the mirror reflects the incident ray according to the law of reflection (∠i = ∠r), and the geometry of the curved surface is such that all these reflected rays pass through the single point F on the principal axis. Hence, all the parallel rays from the Sun converge at the principal focus, forming a bright, intense spot of light on the wall placed at that position.

Question 6

Read the following passage and answer the questions that follow:

A student is setting up an optical bench experiment. She places a concave mirror of focal length 15 cm on the bench. She then places a bright object (a lit candle) at different positions in front of the mirror and records her observations in the table below:

| Trial | Object distance (u) | Nature of image observed |
|-------|--------------------|--------------------------|
| P     | 45 cm              | Real, inverted, diminished |
| Q     | 30 cm              | Real, inverted, same size |
| R     | 20 cm              | Real, inverted, magnified |
| S     | 10 cm              | Virtual, erect, magnified |

The student then replaces the concave mirror with a convex mirror of the same focal length (15 cm) and places the candle at a distance of 30 cm in front of it.

(a) Using the mirror formula, verify whether Trial Q gives the correct image position for the concave mirror. Show your working. [1 mark]

(b) In Trial S, the student claims the image cannot be caught on a screen. Justify her claim using the nature of the image formed. [1 mark]

(c) Calculate the image distance and magnification when the candle is placed 30 cm in front of the convex mirror (f = +15 cm). Hence state the nature and position of the image. [2 marks]

OR

(c) Between Trial P and Trial R (concave mirror), in which trial is the magnitude of magnification greater than 1? Justify your answer using the mirror formula. Also state one practical application that uses the image-forming property demonstrated in Trial R. [2 marks]

Accepted Answer

(a) Verification for Trial Q (concave mirror, u = −30 cm, f = −15 cm):

Using sign convention: u = −30 cm, f = −15 cm

Mirror formula: 1/v + 1/u = 1/f

⟹ 1/v = 1/f − 1/u

⟹ 1/v = 1/(−15) − 1/(−30)

⟹ 1/v = −1/15 + 1/30

⟹ 1/v = (−2 + 1)/30 = −1/30

∴ v = −30 cm

The image is formed at 30 cm in front of the mirror (same side as object), confirming it is real, inverted, and the same size as the object. ✓ Trial Q is verified correct. [1 mark]

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(b) In Trial S, the object is placed between the pole and the focus (u = −10 cm, f = −15 cm). The image formed by a concave mirror when the object is between P and F is virtual and erect — it forms behind the mirror. Since the reflected rays diverge after reflection and only appear to meet behind the mirror, they do not actually converge at any point in front of the mirror. A screen can only capture an image where rays actually meet (real image). Because the rays never actually intersect in front of the mirror, the image cannot be projected onto a screen. [1 mark]

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(c) MAIN OPTION:

For convex mirror: using sign convention, u = −30 cm, f = +15 cm (convex mirror has positive focal length).

Mirror formula: 1/v + 1/u = 1/f

⟹ 1/v = 1/f − 1/u

⟹ 1/v = 1/(+15) − 1/(−30)

⟹ 1/v = 1/15 + 1/30

⟹ 1/v = (2 + 1)/30 = 3/30 = 1/10

∴ v = +10 cm [1 mark]

Magnification: m = −v/u = −(+10)/(−30) = +1/3 ≈ +0.33

Since v is positive → image is formed behind the mirror (virtual).
Since m is positive → image is erect.
Since |m| < 1 → image is diminished.

∴ The image is formed 10 cm behind the convex mirror. It is virtual, erect, and diminished. [1 mark]

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(c) OR OPTION:

In Trial P: u = −45 cm, f = −15 cm.
1/v = 1/(−15) − 1/(−45) = −1/15 + 1/45 = (−3 + 1)/45 = −2/45
∴ v = −22.5 cm
m = −v/u = −(−22.5)/(−45) = −0.5
|m| = 0.5 < 1 → image is diminished.

In Trial R: u = −20 cm, f = −15 cm.
1/v = 1/(−15) − 1/(−20) = −1/15 + 1/20 = (−4 + 3)/60 = −1/60
∴ v = −60 cm
m = −v/u = −(−60)/(−20) = −3
|m| = 3 > 1 → image is magnified. [1 mark]

Therefore, Trial R has magnification of magnitude greater than 1, because the object is placed between F and C (u = −20 cm, which is between f = −15 cm and 2f = −30 cm), producing a real, inverted, magnified image beyond C.

Practical application of Trial R: A concave mirror used as a reflector in a cinema projector or as a doctor's head mirror (or dentist's mirror), where a magnified real image of an illuminated object is required. [1 mark]

Question 7

At a dental clinic in Pune, Maharashtra, Dr. Anjali uses a small concave mirror to examine a patient's teeth. She holds the mirror at a distance of 2 cm from a tooth so that an erect and magnified image of the tooth is formed behind the mirror. The focal length of the dental mirror is 3 cm. During a school visit to the clinic, Class 10 student Aryan observes this procedure and is curious about the optical principles involved.

Accepted Answer

**(a)**

The mirror formula for a spherical mirror is:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

where *v* = image distance, *u* = object distance, and *f* = focal length, all measured from the pole of the mirror.

**(b)**

Since f = R/2, the radius of curvature R = 2f.

$$R = 2 	imes 3 = 6 	ext{ cm}$$

∴ The radius of curvature of the dental mirror is **6 cm**.

**(c)**

Using the New Cartesian Sign Convention: u = −2 cm (object in front of mirror), f = −3 cm (concave mirror).

Applying the mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-3} - \frac{1}{-2} = -\frac{1}{3} + \frac{1}{2} = \frac{-2+3}{6} = \frac{1}{6}$$

$$	herefore v = +6 	ext{ cm}$$

Magnification:

$$m = -\frac{v}{u} = -\frac{(+6)}{(-2)} = +3$$

Since m = +3 is **positive**, the image is **erect and virtual**, formed 6 cm behind the mirror. The magnitude 3 indicates the image is **3 times magnified** compared to the object. This confirms the use of a concave mirror as a dental mirror — when the tooth (object) is placed between the pole and focus, it produces a virtual, erect, and magnified image.

Question 8

A traffic surveillance camera is mounted at a busy intersection. The camera uses a convex mirror as a wide-angle reflector to capture vehicles approaching from a large field of view. The security officer notices that in the mirror's image, all vehicles — whether 5 metres away or 50 metres away — appear smaller than their actual size, and the image always appears to be located close to the mirror. The officer then replaces the convex mirror with a concave mirror of the same radius of curvature, hoping to get a larger image. However, he finds that the concave mirror is not suitable for this purpose.

(a) State the nature (real/virtual), orientation (erect/inverted), and size (magnified/diminished) of the image formed by the convex mirror used in the surveillance system. [1 mark]

(b) The radius of curvature of the convex mirror is 60 cm. A vehicle is 5 m away from the mirror. Using the mirror formula, find the position of the image of the vehicle. [2 marks]

(c) Explain why the concave mirror is NOT suitable as a replacement for the wide-angle surveillance mirror, even though it can form a magnified image under certain conditions.
OR
(c) The officer wants to use a mirror that gives an upright image of a vehicle at all positions of the object. Name the type of mirror and state any TWO properties of the image it always forms, giving a reason for each property. [1 mark]

Accepted Answer

(a) The image formed by the convex mirror is: virtual, erect (upright), and diminished (smaller than the object). [1 mark]

(b) Given:
Radius of curvature, R = 60 cm (convex mirror → R is positive)
Using the relation f = R/2:
f = +60/2 = +30 cm

Object distance, u = −500 cm (object is 5 m = 500 cm in front of mirror; u is always negative by sign convention)

Using mirror formula:
1/v + 1/u = 1/f
⟹ 1/v + 1/(−500) = 1/(+30)
⟹ 1/v = 1/30 + 1/500
⟹ 1/v = 500/(30 × 500) + 30/(30 × 500)
⟹ 1/v = (500 + 30)/15000
⟹ 1/v = 530/15000
⟹ v = 15000/530
⟹ v ≈ +28.3 cm

∴ The image is formed approximately 28.3 cm behind the mirror (positive sign confirms the image is virtual, formed behind the mirror). [2 marks]

(c) The concave mirror is NOT suitable as a wide-angle surveillance mirror for the following reasons:

(i) A concave mirror converges light — it has a limited field of view, capturing only a narrow region in front of it. A surveillance system requires a wide field of view to monitor vehicles from many directions simultaneously. The convex mirror diverges reflected rays and therefore has a much wider field of view than a concave mirror of the same size.

(ii) The image formed by a concave mirror changes its nature, position, and size drastically depending on where the object (vehicle) is placed relative to the focal point. When the vehicle is beyond the centre of curvature, the image is real, inverted, and beyond the mirror — it cannot be seen directly in the mirror. When the vehicle is between F and P, the image is virtual but magnified — showing only a small region. The image is therefore inconsistent and unpredictable for surveillance purposes. The convex mirror always forms a virtual, erect, and diminished image regardless of the object's position, giving a consistent, complete view at all times. [1 mark]

OR

(c) The type of mirror that always gives an upright image at all positions of the object is a CONVEX mirror.

Property 1 — The image is always virtual.
Reason: A convex mirror is a diverging mirror. The reflected rays always diverge after reflection and never actually meet in front of the mirror. They appear to meet behind the mirror when extended backwards. Since the rays do not actually converge, the image cannot be obtained on a screen — it is always virtual.

Property 2 — The image is always diminished (smaller than the object).
Reason: The principal focus and centre of curvature of a convex mirror lie behind the mirror. For any position of the object in front of the mirror, the image is always formed between the pole (P) and the principal focus (F), i.e., within a very small region behind the mirror. This means the image is always compressed into a smaller space than the object occupies, making it diminished for all object positions. [1 mark]

Question 9

The absolute refractive index of kerosene and turpentine oil are 1.44 and 1.47 respectively. If the speed of light in vacuum is 3 × 10⁸ m/s, the speed of light in kerosene is approximately:

Accepted Answer

Option (A) is correct.

Explanation: The absolute refractive index of a medium is defined as n = c/v, where c is the speed of light in vacuum and v is the speed of light in the medium. For kerosene, n = 1.44 and c = 3 × 10⁸ m/s, so v = c/n = (3 × 10⁸)/1.44 ≈ 2.08 × 10⁸ m/s. The refractive index of turpentine oil (1.47) is not required for this calculation.

Question 10

The radius of curvature of a concave mirror is 24 cm. What is its focal length?

Accepted Answer

Option (B) is correct.
Explanation: For any spherical mirror, the focal length f is related to the radius of curvature R by the relation f = R/2. Here, R = 24 cm, so f = 24/2 = 12 cm. Since the mirror is concave, by the New Cartesian Sign Convention the focal length is taken as negative (f = −12 cm), but the magnitude of the focal length is 12 cm.

Question 11

The absolute refractive index of ethanol and ruby are 1.36 and 1.76 respectively. If the speed of light in ethanol is 2.21 × 10⁸ m/s, what is the speed of light in ruby?

Accepted Answer

Option (A) is correct.

Explanation: The absolute refractive index of a medium is defined as n = c/v, where c is the speed of light in vacuum and v is the speed of light in that medium. First, c is calculated using the data for ethanol: c = n_ethanol × v_ethanol = 1.36 × 2.21 × 10⁸ = 3.00 × 10⁸ m/s. Then, applying the same relation to ruby: v_ruby = c / n_ruby = (3.00 × 10⁸) / 1.76 ≈ 1.71 × 10⁸ m/s. Since ruby has a higher refractive index than ethanol (1.76 > 1.36), light travels more slowly in ruby, which is consistent with this result.

Question 12

The absolute refractive indices of diamond and glass are 2.42 and 1.50 respectively. Which of the following statements correctly compares these two media?

Accepted Answer

Option (A) is correct.

Explanation: The absolute refractive index of a medium is defined as n = c/v, where c is the speed of light in vacuum and v is the speed of light in that medium. A higher refractive index indicates greater optical density and a lower speed of light. Since the refractive index of diamond (2.42) is greater than that of glass (1.50), diamond is optically denser than glass. Consequently, the speed of light in diamond (v = c/2.42) is less than the speed of light in glass (v = c/1.50), meaning light travels slower in diamond than in glass.

Question 13

A telescope objective consists of two thin lenses placed in contact. The first lens is a converging lens of focal length 10 cm and the second lens is a diverging lens of focal length 25 cm. What is the net power of this lens combination?

Accepted Answer

Option (A) is correct.

Explanation: The power of a lens is given by P = 1/f, where f must be in metres. For the converging lens, f₁ = +10 cm = +0.10 m, so P₁ = 1/0.10 = +10 D. For the diverging lens, f₂ = −25 cm = −0.25 m (negative because diverging), so P₂ = 1/(−0.25) = −4 D. For lenses placed in contact, the net power is P = P₁ + P₂ = +10 + (−4) = +6 D.

Question 14

When white light passes through a glass prism, it splits into a band of colours. Which of the following correctly identifies the colour that bends the MOST and the colour that bends the LEAST during dispersion?

Accepted Answer

Option (A) is correct.

Explanation: During dispersion of white light through a glass prism, each colour has a different wavelength and therefore a different refractive index in glass. Shorter wavelengths have a higher refractive index and bend more, while longer wavelengths have a lower refractive index and bend less. Violet light has the shortest wavelength among all visible colours and therefore bends the most, while red light has the longest wavelength and therefore bends the least. This produces the spectrum VIBGYOR, with violet at one end and red at the other. Options (B), (C), and (D) are incorrect because (B) reverses the correct order, and (C) and (D) cite intermediate colours that do not represent the extremes of bending.

Question 15

A student uses two thin lenses in contact to make a compound lens system. The first lens has a focal length of +25 cm and the second lens has a focal length of −50 cm. What is the net power of this combination?

Accepted Answer

Option (A) is correct.

Explanation: The power of a lens is given by P = 1/f, where f is the focal length in metres. For the first lens, f₁ = +25 cm = +0.25 m, so P₁ = 1/0.25 = +4 D. For the second lens, f₂ = −50 cm = −0.50 m, so P₂ = 1/(−0.50) = −2 D. For two thin lenses in contact, the net power is P = P₁ + P₂ = +4 + (−2) = +2 D.

Question 16

The power of a lens is +2.0 D. What type of lens is it and what is its focal length?

Accepted Answer

Option (B) is correct.
Explanation: The power of a lens is given by P = 1/f, where f is the focal length in metres. Since P = +2.0 D, the focal length f = 1/P = 1/2.0 = 0.5 m = 50 cm. A positive power indicates a converging (convex) lens. Therefore, the lens is a convex lens with focal length 50 cm.

Question 17

The absolute refractive index of glycerine and flint glass are 1.47 and 1.65 respectively. If the speed of light in glycerine is 2.04 × 10⁸ m/s, what is the speed of light in flint glass?

Accepted Answer

Option (A) is correct.

Explanation: The absolute refractive index of a medium is defined as n = c/v, where c is the speed of light in vacuum and v is the speed of light in that medium. First, the speed of light in vacuum is found using the data for glycerine: c = n_glycerine × v_glycerine = 1.47 × 2.04 × 10⁸ = 3.0 × 10⁸ m/s. The speed of light in flint glass is then v_flint = c/n_flint = (3.0 × 10⁸)/1.65 = 1.82 × 10⁸ m/s, which matches Option (A).

Question 18

An object is placed 30 cm in front of a convex mirror of focal length 15 cm. The image formed is:

Accepted Answer

Option (B) is correct.

Explanation: A convex mirror always forms a virtual, erect and diminished image, regardless of the position of the object in front of it. This can be confirmed using the mirror formula. Using sign convention, u = −30 cm and f = +15 cm (focal length of a convex mirror is positive). Applying 1/v + 1/u = 1/f gives 1/v = 1/15 − 1/30 = (2 − 1)/30 = 1/30, so v = +30 cm... wait — recalculating: 1/v = 1/f − 1/u = 1/15 − 1/(−30) = 1/15 + 1/30 = 2/30 + 1/30 = 3/30 = 1/10, so v = +10 cm. Since v is positive, the image is formed behind the mirror, confirming it is virtual and erect. Magnification m = −v/u = −(+10)/(−30) = +1/3, which is positive (erect) and less than 1 (diminished). Hence the image is virtual, erect and diminished.

Question 19

A convex lens has a focal length of 25 cm. What is the power of this lens?

Accepted Answer

Option (A) is correct.
Explanation: The power of a lens is given by P = 1/f, where f must be expressed in metres. A convex lens has a positive focal length, so f = +25 cm = +0.25 m. Therefore, P = 1/0.25 = +4 D. A convex (converging) lens always has positive power, which rules out options (B) and (D). Option (C) is incorrect because 25 D would require a focal length of only 4 cm, not 25 cm.

Question 20

A ray of light travelling parallel to the principal axis strikes a concave mirror. After reflection, through which point does this ray pass?

Accepted Answer

Option (B) is correct.

Explanation: For a concave mirror, any ray of light travelling parallel to the principal axis, after reflection, passes through the principal focus (F). This is the defining property of the principal focus of a concave mirror — it is the point on the principal axis where all rays parallel to the principal axis converge after reflection.

Light - Reflection and Refraction: Class 10 Science Practice Questions

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