# Light reflection and refraction Worksheet-2

Light reflection and refraction Worksheet-2

1. Light travels through water with s speed of 2.25 × 108 m/s. what is the refractive index of water? Given speed of light in vacuum = 3 × 108 m/s.

A. 1.33                B. 0.75                C. 0.707             D. 1.5

1. Light travels from a rarer medium 1 to a denser medium 2. the angle of incidence and refraction are respectively 45° and 30°. Calculate the refractive index of second medium with respect to the first medium.

A. /2                                          B.

C. 1/                                           D. None of these

1. Light travels from a rarer medium 1 to a denser medium 2. The angle of incidence and refraction are respectively 45° and 30°. What is the refractive index of medium 1 w.r.t. medium 2 ?

A.                   B. /2              C. 2/              D. 2

1. A pond of depth 20 cm is filled with water of refractive index 4/3. Calculate apparent depth of the tank when viewed normally.

A. 10 cm             B. 15 cm             C. 5 cm               D. 26.67 cm

1. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is cm 15 cm. Find the nature, position and size of the image.

A. real and inverted, v = 30 cm, h2 = –4.0 cm

B. Virtual and inverted, v = 30 cm, h2 = 4.0 cm

C. real and inverted, v = 15 cm, h2 = 4.0 cm

D. None of these

1. A

Here, speed of light in water, u = 2.25 × 108 m/s

Speed of light in vacuum, c = 3 × 108 m/s, refractive index n = ?

From

1. B

Here, angle of incidence i = 45° and angle of refraction, r = 30°

refractive index of medium 2 w.r.t. medium 1n2 = ?

According to Snell’s law,

1. B

Here, 2n1 = ?

As

1. B

Here, real depth of pond x = 20 cm

Refractive index of water

Apparent depth, y = ?

As

1. A

Given object size, h1 = 2.0 cm

Focal length of convex lens, f = 10 cm

Object distance, u = –15 cm

Image distance, v = ?

Image size h2 = ?

According to Lens formula

∴

v = 30 cm

As v is positive, the image formed is on the right side of the lens. It must be real and inverted. Now

Linear magnification,

∴

h2 = –4.0 cm

And  m = –2

Negative sign of m and h2 show that the image is inverted.

Thus a real, inverted image enlarged 2 times (i.e. 4.0 cm tall) is formed at a distance of 30 cm on the right side of the lens.