NCERT Class 10 Science: Light Reflection and Refraction

Reflection of Light and Spherical Mirrors

Reflection is the phenomenon where light bounces back into the same medium after striking a polished surface. For a plane mirror, the laws of reflection state that the angle of incidence equals the angle of reflection, and the incident ray, normal, and reflected ray all lie in the same plane.

In class 10 science light reflection and refraction, we focus deeply on spherical mirrors. These are mirrors whose reflecting surfaces are parts of a sphere:

• Concave Mirror: Reflecting surface is curved inwards. It is a converging mirror and can form both real (inverted) and virtual (erect) images depending on the object's position.
• Convex Mirror: Reflecting surface is curved outwards. It is a diverging mirror and always forms a virtual, erect, and diminished image.

To calculate image positions, we use the Mirror Formula:
1/f = 1/v + 1/u
Where f is the focal length, v is the image distance, and u is the object distance. Magnification (m) is given by the ratio of image height to object height: m = h'/h = -v/u.

Refraction of Light & Snell's Law

1. Bending of Light Rays — Refraction is the bending of light when it passes obliquely from one transparent medium to another. When light goes from a rarer medium (like air) to a denser medium (like glass), it bends towards the normal. Conversely, it bends away from the normal when traveling from a denser to a rarer medium.
2. Applying Snell's Law — The second law of refraction, known as Snell's Law, states that the ratio of the sine of the angle of incidence (i) to the sine of the angle of refraction (r) is constant for a given pair of media: sin(i) / sin(r) = constant = n. This constant is the refractive index of the second medium with respect to the first.
3. Calculating Absolute Refractive Index — The absolute refractive index (n_m) of a medium is defined as the ratio of the speed of light in air/vacuum (c = 3 * 10^8 m/s) to the speed of light in that specific medium (v). The formula is: n_m = c / v. Since it is a ratio of identical quantities, it has no units.

Solved Numerical Examples (Mirror & Lens)

• Example 1 (Convex Mirror): An object is placed 10 cm in front of a convex mirror of focal length 15 cm. Find the image position. - Step 1: Identify given values with Cartesian signs. For a convex mirror, focal length f = +15 cm, and object distance u = -10 cm. - Step 2: Write down the Mirror Formula: 1/f = 1/v + 1/u. - Step 3: Rearrange to find 1/v: 1/v = 1/f - 1/u. - Step 4: Substitute values: 1/v = 1/15 - (1/-10) = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6. - Step 5: Solve for v: v = +6 cm. The positive sign indicates that the image is formed virtual and erect, 6 cm behind the mirror.
• Example 2 (Concave Lens): A concave lens has a focal length of 15 cm. At what distance should an object be placed so that it forms an image at 10 cm from the lens? - Step 1: Identify values. For a concave lens, focal length f = -15 cm and image distance v = -10 cm (since concave lenses always form virtual images on the same side as the object). - Step 2: Use Lens Formula: 1/f = 1/v - 1/u. - Step 3: Rearrange for 1/u: 1/u = 1/v - 1/f. - Step 4: Substitute values: 1/u = 1/-10 - (1/-15) = -1/10 + 1/15 = (-3 + 2)/30 = -1/30. - Step 5: Solve for u: u = -30 cm. The object must be placed at a distance of 30 cm in front of the lens.

Board Exam Secret: The New Cartesian Sign Convention

Over 80% of students lose numerical marks in cbse class 10 light reflection and refraction due to simple sign errors! Follow these golden rules to avoid mistakes on your sketchpad:

1. Object Distance (u) is ALWAYS Negative: The object is always placed to the left of the mirror or lens, so u is always negative.
2. Focal Length (f):

• For Concave mirror/lens: f is always negative.
• For Convex mirror/lens: f is always positive.

1. Real vs. Virtual Images:

• A negative image distance (v) for a lens means virtual; for a mirror, it means real.
• A positive image distance (v) for a lens means real; for a mirror, it means virtual.

1. Magnification (m): If m is negative, the image is real and inverted. If m is positive, the image is virtual and erect.

Practice Questions with Solutions

• Q: A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located? A: Step 1: Write down the given values. Object distance, u = -10 cm. Step 2: Real image implies magnification m is negative. Therefore, m = -3. Step 3: Use the magnification formula for mirrors: m = -v/u. Step 4: Substitute values: -3 = -v / (-10). Step 5: Solve for v: -3 = v / 10 which gives v = -30 cm. Final answer: The image is located at a distance of 30 cm in front of the concave mirror (on the same side as the object).
• Q: Light travels from air into a medium of refractive index 1.5. Calculate the speed of light in this medium. (Speed of light in air = 3 10^8 m/s). A: Step 1: Write down the formula for refractive index: n = c / v. Step 2: Identify given values: n = 1.5 and c = 3 10^8 m/s. Step 3: Rearrange the formula to solve for velocity in the medium: v = c / n. Step 4: Substitute the values: v = (3 10^8) / 1.5. Step 5: Calculate: v = 2 10^8 m/s. Final answer: The speed of light in the medium is 2 * 10^8 m/s.
• Q: A convex lens of focal length 10 cm is placed at some distance from a candle. If the image is formed on a screen placed 20 cm on the other side of the lens, find the position of the candle. A: Step 1: Identify given values. For a convex lens forming an image on a screen (real image), focal length f = +10 cm, image distance v = +20 cm. Step 2: Write down the Lens Formula: 1/f = 1/v - 1/u. Step 3: Rearrange for 1/u: 1/u = 1/v - 1/f. Step 4: Substitute values: 1/u = 1/20 - 1/10 = (1 - 2)/20 = -1/20. Step 5: Solve for u: u = -20 cm. Final answer: The candle is placed 20 cm in front of the convex lens.
• Q: Find the focal length of a corrective lens of power -2.0 D. What type of lens is this? A: Step 1: Write down the relation between Power (P) and Focal length (f in meters): P = 1/f. Step 2: Substitute the power value: -2.0 = 1/f. Step 3: Solve for f: f = -1 / 2.0 = -0.5 m. Step 4: Convert to centimeters: f = -50 cm. Step 5: Determine the type of lens based on the sign. Since focal length is negative, it is a diverging (concave) lens. Final answer: The focal length of the corrective lens is -0.5 m (or -50 cm), and it is a concave lens.

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