This chapter explores how light interacts with surfaces and how its path can be bent:

**Reflection:**

**Bouncing Back:**Reflection is the phenomenon where light bounces back into the same medium when it strikes a smooth surface like a mirror.**Key Concepts:****Incident Ray:**Light ray falling on the surface.**Reflected Ray:**Light ray bouncing back from the surface.**Normal:**A line drawn perpendicular to the reflecting surface at the point of incidence.**Angle of Incidence:**Angle between the incident ray and the normal.**Angle of Reflection:**Angle between the reflected ray and the normal.**Laws of Reflection:**- The angle of incidence is always equal to the angle of reflection. (i = r)
- The incident ray, the reflected ray, and the normal all lie in the same plane.

**Types of Reflection:****Specular Reflection:**Smooth surfaces produce sharp, well-defined reflected rays (e.g., mirrors).**Diffuse Reflection:**Rough surfaces scatter the reflected rays in various directions (e.g., painted walls).

**Refraction:**

**Bending the Path:**Refraction is the phenomenon where light bends as it travels from one medium (e.g., air) to another (e.g., water) with a different density.**Key Concepts:****Refraction at Interface:**Bending of light occurs at the boundary between two mediums.**Refraction and Speed:**Light generally slows down when entering a denser medium.**Apparent Depth:**Refraction can cause objects partly submerged in water to appear shallower than they actually are.

**Applications:**

- Understanding reflection and refraction is crucial for various technologies like microscopes, telescopes, lenses in eyeglasses, and optical fibers used for communication.

**Questions (Page 142)**

**1. Define the principal focus of a concave mirror.**

**Ans : **The principal focus of a concave mirror, also known as the focal point, is a specific point on its principal axis where parallel rays of light coming in **converge** (come together) after reflecting from the mirror.

**2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?**

**Ans : **

For a spherical mirror, the focal length (f) is half the radius of curvature (R).

Therefore, if the radius of curvature (R) is 20 cm, the focal length (f) can be calculated as:

f = R / 2 f = 20 cm / 2 f = 10 cm

So, the focal length of the spherical mirror is 10 cm.

**3. Name a mirror that can give an erect and enlarged image of an object.**

**Ans : **A concave mirror can give an erect and enlarged image of an object under certain conditions.

**4. Why do we prefer a convex mirror as a rear-view mirror in vehicles ?**

**Ans : **Convex mirrors are king for car rear-view mirrors because of 2 reasons:

**Wider View:**Bulging outwards, they capture more traffic behind you compared to flat mirrors. Imagine a flat mirror showing a narrow hallway – convex mirrors show a wider space, like a panoramic view.**Fewer Blind Spots:**Those annoying areas you can’t see directly? Convex mirrors help reduce them by showing you what a flat mirror wouldn’t. This is crucial for safe driving!

**Questions (Page 145)**

**1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.**

**Ans : **Contrary to a concave mirror, the focal length (f) of a convex mirror isn’t simply half its radius of curvature (R). For convex mirrors, the focal length is **half the radius of curvature with a negative sign**.

Here’s the formula:

f = -R / 2

Given the radius of curvature (R) is 32 cm:

f = – (32 cm) / 2 f = -16 cm

**2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?**

**Ans : **

**Given information:**

- Magnification (m) = -3 (negative because the image is real)
- Object distance (u) = -10 cm (negative because the object is in front of the mirror)

**Formula:**

The mirror formula for a concave mirror relates the object distance (u), image distance (v), and focal length (f) as:

1/u + 1/v = 1/f

m = -v / u

Since we know m = -3 and u = -10 cm, we can solve for v:

v = m * u = (-3) * (-10 cm) = 30 cm

**Questions (Page 150)**

**1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?**

**Ans : **When a ray of light travels obliquely from air into water, the light ray bends **towards the normal**. This happens because water is optically denser than air.

**2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3 x 10****8**** ms****-1****.**

**Ans : **

**Given information:**

- Refractive index of glass (n) = 1.50
- Speed of light in vacuum (c) = 3 x 10^8 m/s (given)

v = c / n

v = (3 x 10^8 m/s) / 1.50

v = 2.00 x 10^8 m/s

**3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.**

**Ans :**

Material | Refractive Index |

Air | 1.0003 |

Water | 1.333 |

Diamond | 2.417 |

**Highest Optical Density:** The material with the **highest refractive index** will have the highest optical density.

**Lowest Optical Density:** The material with the **lowest refractive index** will have the lowest optical density.

Diamond (2.417) has the highest refractive index, indicating the highest optical density.

Air (1.0003) has the lowest refractive index, indicating the lowest optical density.

**4. You are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table 10.3.**

**Ans : **

For kerosene, n = 1.44

For turpentine, n = 1.47

For water, n = 1.33

Based on the relationship between refractive index and light speed, light will travel **fastest in kerosene**

**5. The refractive index of diamond is 2.42. What is the meaning of this statement?**

**Ans : **A diamond’s refractive index of 2.42 tells us that light undergoes a significant change in direction and speed when it enters or exits a diamond.

**Questions (Page 158)**

**1. Define 1 dioptre of power of a lens.**

**Ans : **One dioptre (written as 1 D) is the unit of measurement for the optical power of a lens or curved mirror. It tells you how much the lens or mirror can bend light rays.

**2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens. , Sol. Here, u — +50 cm ..**

**Ans : **

1/u + 1/v = 1/f

We can rearrange this formula to solve for the focal length (f):

f = 1 / (1/u + 1/v)

Plugging in the values we know (u = +50 cm and v = +50 cm):

f = 1 / (1/50 cm + 1/50 cm) f = 1 / (2/50 cm) f = 25 cm

Now that we have the focal length (f), we can find the lens power (P) using the formula:

P = 1 / f

P = 1 / 25 cm (convert cm to meters for the unit of power) P = 1 / 0.25 m P = 4 D (dioptres)

**Therefore:**

- The needle is placed
**50 cm in front of the convex lens**. - The power of the lens is
**4 dioptres (4 D)**.

**3. Find the power of a concave lens of focal length 2 m.**

**Ans : **

**Formula:** The formula for lens power (P) is:

P = 1 / f

**Calculate Power:**

P = 1 / (-2 m) ** (Remember to include the negative sign) **

P = -0.5 D (dioptres)

**Exercises**

**1. Which one of the following materials cannot be used to make a lens ?**

**(a) Water**

**(b) Glass**

**(c) Plastic**

**(d) Clay**

**Ans : **(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) Beyond the centre of curvature**

**(d) Between the pole of the mirror and its principal focus.**

**Ans : **(d) Between the pole of the mirror and its 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.**

**Ans : **(d) Between the optical centre of the lens and its principal focus.

**4. A spherical mirror and a thin spherical lens have each a focal length of -15 cm. 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, but the lens is concave.**

**Ans : **(a) Both concave

**5. No matter how far you stand from mirror, your image appears erect. The mirror is likely to be**

**(a) plane**

**(b) concave**

**(c) convex**

**(d) either plane or convex.**

**Ans : **(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 50 cm.**

**(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.**

**Ans : **(c) A convex lens of focal length 5 cm.

**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.**

**Ans : **

To obtain an erect image with a concave mirror, the object must be placed between the mirror’s pole (P) and its focal point (F).

The focal length (f) of the mirror is given as 15 cm in this case.

So, the object distance (u) must lie in the range of 0 cm to 15 cm (excluding 15 cm) for an erect image.

**Ray Diagram**

- A ray diagram helps visualize how light rays reflect off the mirror to form an image.

- Draw a concave mirror with a labeled pole (P) and focal point (F).
- Place the object (O) between the pole (P) and the focal point (F).
- Draw two rays originating from the top of the object (O):
- Ray 1 parallel to the principal axis.
- Ray 2 heading towards the focal point (F).

- Reflect ray 1 off the mirror, making it bend towards the focal point (F) after reflection.
- Reflect ray 2 parallel to the principal axis after it bounces off the mirror.
- The reflected rays (dashed lines) appear to converge at a point (I) behind the mirror. This point (I) represents the virtual, erect, and magnified image of the object (O).

**8. Name the type of mirror used in the following situations.**

**(a) Headlights of a car.**

**(b) Side/rear-view mirror of a vehicle.**

**(c) Solar furnace.**

**Support your answer with reason.**

**Ans : **

**(a) Headlights of a car:** A **concave mirror** is used in car headlights.

**Reasoning:**Concave mirrors have the property of converging light rays to a focal point. In car headlights, a bulb or LED is placed near the focal point of the concave mirror. This causes the reflected light rays to travel almost parallel, forming a powerful beam that illuminates the road ahead.

**(b) Side/rear-view mirror of a vehicle:** A **convex mirror** is used in car side/rear-view mirrors.

**Reasoning:**Convex mirrors bulge outwards, providing a wider field of view compared to flat mirrors. This is crucial for drivers to see more traffic behind them than they could with a flat mirror. While the image is slightly shrunk, the wider view enhances safety by reducing blind spots.

**(c) Solar furnace:** A **concave mirror** is used in a solar furnace.

**Reasoning:**Similar to car headlights, a large concave mirror is used in a solar furnace. Sunlight strikes the mirror’s surface and gets reflected inwards, converging on a focal point. This concentrated solar energy creates a very high temperature at the focal point, which can be used for various purposes like melting materials or generating heat.

**9. 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.**

**Ans : **

**Objective:** Determine if a convex lens covered with black paper on one half can produce a complete image of an object.

**Materials:**

- Convex lens
- Black paper
- Light source (candle or flashlight)
- White screen or wall

**Procedure:**

- Place the convex lens on a table.
- Light the candle or turn on the flashlight.
- Position the white screen or wall a distance greater than the focal length of the lens (typically a few tens of centimeters).
- Observe the image formed on the screen when the entire lens is uncovered. Note its characteristics (brightness, completeness).
- Take a piece of black paper and carefully cover one half of the convex lens.
- While keeping the setup unchanged, observe the image formed on the screen now.
- Record your observations about the image (presence, completeness, brightness compared to step 4).

**Observations:**

You should observe the following:

**Step 4:**With the entire lens uncovered, a clear, inverted, and real image of the light source (candle flame or bulb filament) should be visible on the screen.**Step 7:**Even with half the lens covered, a complete image of the light source should still be visible on the screen. However, the image might appear slightly fainter compared to the complete lens case.

**Explanation:**

A convex lens converges light rays passing through it to form a focused image. Each half of the convex lens acts like a separate lens with its own focal point. When half the lens is covered, light rays are still able to pass through the uncovered half and converge to form an image.

**10. 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.**

**Ans : **

Here : Object distance, u= -25 cm,

Object height, h = 5 cm,

Focal length, f = +10 cm

According to the lens formula,

According to the lens formula, 1f=1ν−1u , we have

⇒ 1ν=1f−1u=110−125=15250orν=25015

=16.66cm

The positive value of v shows that the image is formed at the other side of the lens.

The negative value of image height indicates that the image formed is inverted.

The position, size, and nature of image are shown alongside in the ray diagram.

**11. 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.**

**Ans : **Focal length, f = -15 cm, Image distance, ν = -10 cm (as concave lens forms the image on the same side of the lens)

From the lens formula

1f=1ν−1u , we have

Object distance, u = -30 cm

The negative value of u indicates that the object is placed in front of the lens.

**12. 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.**

**Ans : **

**Formula:**

We can use the mirror formula to determine the image position (v) for a spherical mirror:

1/u + 1/v = 1/f

where:

- u = object distance (positive value as it’s in front of the mirror) = 10 cm
- v = image distance (to be determined)
- f = focal length of the mirror (positive for convex mirrors) = 15 cm

**Calculation:**

- Plug the values into the formula:

1/10 + 1/v = 1/15

- Solve for v:

- Take the reciprocal of both sides:

v = 1 / (1/10 + 1/15)

- Find a common denominator for the fractions:

v = 1 / (3 + 2) / 30

v = 1 / (5/30)

v = 6 cm

**13. The magnification produced by a plane mirror is +1. What does this mean ?**

**Ans : **m = h’ / h

m = +1 = h’ / h

Therefore, h’ = h, indicating the image size (h’) is the same as the object size (h).

A magnification of +1 for a plane mirror tells you that the image reflects the object with its original size and upright orientation.

**14. 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.**

**Ans : **1/u + 1/v = 1/f

where:

- u = object distance (positive value as it’s in front of the mirror) = 20 cm
- v = image distance (to be determined)
- f = focal length of the mirror (positive for convex mirrors) = 15 cm

**Calculation:**

- Plug the values into the formula:

1/20 + 1/v = 1/15

- Solve for v:

- Take the reciprocal of both sides:

v = 1 / (1/20 + 1/15)

- Find a common denominator for the fractions:

v = 1 / (3 + 4) / 60

v = 1 / (7/60)

v = -10 cm (negative sign indicates the image is behind the mirror)

**15. 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 focussed image can be obtained ? Find the size and the nature of the image.**

**Ans : **

We can use the mirror formula to determine the image distance (v) for a spherical mirror:

1/u + 1/v = 1/f

where:

- u = object distance (positive value as it’s in front of the mirror) = 27 cm
- v = image distance (to be determined)
- f = focal length of the mirror (negative for concave mirrors) = -18 cm (convention)

**Calculation:**

- Plug the values into the formula:

1/27 + 1/v = 1/-18

- Solve for v:

- Take the reciprocal of both sides:

v = 1 / (1/27 – 1/18)

- Find a common denominator for the fractions:

v = 1 / (-1 + 3) / 54

v = 1 / (2/54)

v = -54 cm (negative sign indicates the image is formed behind the mirror)

**Therefore, the screen needs to be placed 54 cm away from the mirror (behind it) to obtain a sharp focused image.**

**Image Nature:**

Since the image distance (v) is negative and the object distance (u) is positive for a concave mirror in this case, the image is:

**Real:**Light rays converge to a point at the image location, forming a real image.**Inverted:**The image will have an opposite orientation compared to the object.

**Image Size (h’):**

We can use the magnification formula to find the relative size (h’) of the image compared to the object size (h):

m = h’/h = -v/u

where:

- m = magnification
- h’ = image size
- h = object size (7.0 cm)
- v = image distance (-54 cm)
- u = object distance (27 cm)

**Calculation:**

- Plug the values into the formula:

m = h’/7.0 cm = – (-54 cm) / (27 cm)

- Solve for m:

m = 2

**Since the magnification (m) is 2, the image size (h’) will be twice the object size (h):**

h’ = m * h = 2 * 7.0 cm = 14.0 cm

**16. Find the focal length of a lens of power -2.0 D. What type of lens is this ?**

**Ans : **

**f (in meters) = 1 / D**

where:

- f is the focal length in meters
- D is the power of the lens in dioptres (D)

**Calculation:**

- D = -2.0 D (given)
- f = 1 / (-2.0 D)
**f = -0.5 meters**

**17. 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 ?**

**Ans : **

where:

- f is the focal length in meters
- D is the power of the lens in dioptres (D)

**Given:**

- D = +1.5 D (positive value)

**Calculation:**

- Substitute the value of D into the formula:

f = 1 / (+1.5 D)

- Calculate the focal length:

f = 0.67 meters (positive value)