10th Class Physics Unit 14 Light Numericals with Solutions

14.1.The speed of light in water is 2.25 × 10^8 m s^-1. Calculate the refractive index of water.

Given

Speed of light in vacuum c = 3 × 10⁸ m s^-1

Speed of light in water v = 2.25 × 10⁸ m s^-1

To find

Refractive index of water n = ?

Formula

Using the formula for refractive index n = cv

Working

Substituting the values n = 3 × 10⁸2.25 × 10⁸

Simplifying n = 32.25 = 1.333

Result

The refractive index of water is n ≈ 1.33

14.2.A light ray in air (n₁ = 1.0) strikes a glass surface (n₂ = 1.5) at an angle of incidence 30°. Find the angle of refraction inside the glass.

Given

Refractive index of air n₁ = 1.0

Refractive index of glass n₂ = 1.5

Angle of incidence i = 30°

To find

Angle of refraction r = ?

Formula

Using Snell's Law n₁ sin i = n₂ sin r

Working

Substituting the values 1.0 × sin 30° = 1.5 × sin r

Solving for sin r sin r = 0.51.5 = 13

Computing sin r sin r ≈ 0.3333

Finding the angle r = sin^-1(0.3333)

Result

The angle of refraction is r ≈ 19.47° or 19.5°

14.3.A beam of light in air hits a glass slab at an incidence angle of 40°. Find the angle of refraction inside the glass.

Given

Refractive index of air n₁ ≈ 1.0

Refractive index of glass n₂ ≈ 1.52

Angle of incidence i = 40°

To find

Angle of refraction r = ?

Formula

Using Snell's Law n₁ sin i = n₂ sin r

Working

Putting the values 1.0 × sin 40° = 1.52 × sin r

Substituting sin 40° 0.6428 ≈ 1.52 sin r

Solving for sin r sin r = 0.64281.52 ≈ 0.4229

Finding the angle r = sin^-1(0.4229)

Result

The angle of refraction is r ≈ 25.0°

14.4.A ray of light travels from air into an unknown liquid. The angle of incidence in air is 45°, and the angle of refraction in the liquid is 30°. Calculate the refractive index of the liquid.

Given

Refractive index of air n₁ = 1.0

Angle of incidence i = 45°

Angle of refraction r = 30°

To find

Refractive index of liquid n₂ = ?

Formula

Using Snell's Law n₁ sin i = n₂ sin r

Working

Putting the values 1.0 × sin 45° = n₂ × sin 30°

Substituting values 0.7071 ≈ n₂ × 0.5

Solving for n₂ n₂ = 0.70710.5 = 1.4142

Result

The refractive index of liquid is n₂ ≈ 1.41

14.5.A ray of light passes from air into a liquid. The angle of incidence in air is 50°, and the angle of refraction inside the liquid is 32°. The speed of sound in air is 340 m s^-1 and in the same liquid is 1500 m s^-1. Calculate: (a) the refractive index of the liquid (b) the speed of light in the liquid.

Given

Refractive index of air n₁ = 1.0

Angle of incidence i = 50°

Angle of refraction r = 32°

Speed of light in vacuum c = 3 × 10⁸ m s^-1

To find

(a) Refractive index of liquid n₂ = ?

(b) Speed of light in liquid v = ?

Formula

(a) Using Snell's Law n₁ sin i = n₂ sin r

Working

Putting the values 1.0 × sin 50° = n₂ × sin 32°

Substituting values 0.7660 ≈ n₂ × 0.5299

Solving for n₂ n₂ = 0.76600.5299 = 1.445

Formula

(b) Using the refractive index formula n₂ = cv

Rearranging v = cn₂ = 3 × 10⁸1.445

Computing speed of light in liquid v ≈ 2.076 × 10⁸ m s^-1

Result

The refractive index of liquid is n₂ ≈ 1.445 and speed of light in liquid is v ≈ 2.08 × 10⁸ m s^-1

14.6.Light travels inside a glass block of refractive index n = 1.6. it strikes the glass-air boundary. Calculate the critical angle for total internal reflection.

Given

Refractive index of glass n₁ = 1.6

Refractive index of air n₂ = 1.0

To find

Critical angle C = ?

Formula

Using the formula for critical angle when going from denser medium to air sin C = n₂n₁

Working

Putting the values sin C = 1.01.6 = 0.625

Finding the angle C = sin^-1(0.625)

Result

Critical angle is C ≈ 38.68°

14.7.An object is placed 15.0 cm in front of a convex mirror, forming a virtual image 7.5 cm behind the mirror. What is the focal length of the mirror?

Given

Object distance p = 15 cm

Image distance (negative for virtual image) q = -7.5 cm

To find

Focal length f = ?

Formula

Using the mirror formula 1f = 1p + 1q

Working

Putting the values 1f = 115 + 1-7.5

Computing fractions 1f = 0.0667 - 0.133 = -0.066

Finding focal length f = 1-0.0666 ≈ -15.0 cm

Result

Focal length of mirror is f = -15 cm

14.8.An object 30 cm tall is located 10.5 cm from a concave mirror with focal length 16 cm. (a) where is the image located? (b) how high is it?

Given

Object height h_o = 30 cm

Object distance p = 10.5 cm

Focal length f = 16 cm

To find

(a) Image distance q = ?

(b) Image height h_i = ?

Formula

(a) Using the mirror formula 1f = 1p + 1q

Working

Putting the values 116 = 110.5 + 1q

Rearranging 1q = 116 - 110.5

Computing fractions 1q = 0.0625 - 0.09524 = -0.03274

Finding image distance q = 1-0.03274 ≈ -30.8554 cm

Formula

(b) Using magnification formula Magnification = M = h_ih_o = -qp

Working

Putting the values M = -(-30.54)10.5 = 30.5410.5 ≈ 2.908

Finding image height h_i = M × h_o = 2.908 × 30 ≈ 87.24 cm

Result

(a) Image distance is q ≈ -30.6 cm (virtual image, behind mirror)

(b) Image height is h_i ≈ 87.3 cm (upright, enlarged)

14.9.An object is placed at 30 cm to the left of a concave lens with focal length 15 cm. find the image distance and describe the image.

Given

Object distance p = 30 cm

Focal length of concave lens f = -15 cm

To find

Image distance q = ?

Formula

Using the lens formula 1f = 1p + 1q

Working

Putting the values 1-15 = 130 + 1q

Rearranging 1q = 1-15 - 130

Finding common denominator 1q = -115 - 130 = -330

Simplifying 1q = -110

Result

The image distance is q = -10 cm

Image nature The negative sign shows image is virtual and formed on same side of the object

Unit 14: Light — Numericals