Ray Optics

Three observes A, B, C measure speed of light from a source in vacuum to be vA , vB , vC respectively.

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Q: Three observes A, B, C measure speed of light from a source in vacuum to be vA , vB , vC respectively . If A was moving towards source, B was moving away from source with same speed and C was at rest, then

(a) vA < vB < vC

(b) vA < vB < vC

(c) vC = (vB + vA)/2

(d) vA = vB = vC

Click to See Answer :
Ans: (c) , (d)
Sol: Speed of light in vacuum is an absolute constant. It does not depend upon velocity of observer.
As vA = vB = vC , therefore, vC = (vB + vA)/2

 

Photographs of the ground are taken from an aircraft flying at an altitude of 3 km by a camera with a lens of focal length 50 cm…..

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Q: Photographs of the ground are taken from an aircraft flying at an altitude of 3 km by a camera with a lens of focal length 50 cm. The size of the film in the camera is 18 cm × 18 cm. What is the area of the ground that can be photographed by this camera ?

(a) 720 m × 720 m2

(b) 240 m × 240 m2

(c) 1080 m × 1080 m2

(d) 100 m × 100 m2

Click to See Answer :
Ans: (c)
Sol: As ground is at very large distance from camera, therefore image of ground can be assumed to be formed at the focus of the camera lens.

v = 50 m = 0.5 m , u =-3 km = -3000 m

If x is length of ground photographed, then

$\displaystyle m = |\frac{h_i}{h_o}| = |\frac{v}{u}|$

$\displaystyle |\frac{0.18}{x}| = |\frac{0.5}{3000}|$

x = 1080 m

Area of ground photographed = 1080 m × 1080 m

 

A beautiful person with two normal eyes wants to see full width of her face by a plane mirror….

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Q: A beautiful person with two normal eyes wants to see full width of her face by a plane mirror. The eye to eye and ear to ear distance of her face are 10 cm and 14 cm respectively The minimum width of required mirror is :

Sol: Here, eye to eye distance, a=10 cm ear to ear distance, d = 14 cm
Minimum width of required mirror is

$\displaystyle x = \frac{b-a}{2}$

$\displaystyle x = \frac{14-10}{2}$

x = 2 cm

Diameter of a Plano – convex lens is 6 cm and thickness at the center is 3 mm , as shown in figure ….

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Q: Diameter of a Plano – convex lens is 6 cm and thickness at the center is 3 mm , as shown in figure . If speed of light in material of lens is 2 × 108 m/s, the focal length of the lens is :

Numerical

(a) 15 cm

(b) 20 cm

(c) 30 cm

(d) 10 cm

Click to See Answer :
Ans: (c)

Sol: Here, d = 6 cm, h = 3 mm = 0.3 cm,

v = 2 × 108 m/s , f = ?

If R is radius of curvature of the convex surface of lens, then from

$\displaystyle (\frac{d}{2})^2 = (2R – h)h $

$\displaystyle (\frac{d}{2})^2 = 2 R h $ (Since h << R)

$\displaystyle R = \frac{d^2}{8 h} $

$\displaystyle R = \frac{6^2}{8 \times 0.3} $

$\displaystyle R = 15 cm $

$\displaystyle \mu = \frac{c}{v} $

$\displaystyle \mu = \frac{3 \times 10^8}{2 \times 10^8} $

μ = 1.5

$\displaystyle \frac{1}{f} = (\mu -1)(\frac{1}{R_1} -\frac{1}{R_2})$

$\displaystyle \frac{1}{f} = (1.5 -1)(\frac{1}{\infty} -\frac{1}{-15})$

f = 30 cm

 

A Plano convex lens fits exactly into a Plano concave lens. Their plane surface are parallel to each other….

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Q: A Plano convex lens fits exactly into a Plano concave lens. Their plane surface are parallel to each other. If lenses are made of different materials of refractive indices μ1 and μ2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of combination is

$\displaystyle (a) \frac{2 R}{\mu_2 – \mu_1} $

$\displaystyle (b) \frac{R}{2 (\mu_2 – \mu_1)} $

$\displaystyle (c) \frac{R}{2 (\mu_1 – \mu_2)} $

$\displaystyle (d) \frac{R}{\mu_1 – \mu_2} $

Click to See Answer :
Ans: (d)

Sol: Using Lens Maker’s formula ,

Numerical

$\displaystyle \frac{1}{f_1} = (\mu_1 – 1)( \frac{1}{\infty} – \frac{1}{-R} )$

$\displaystyle \frac{1}{f_1} = \frac{\mu_1 – 1}{R} $

Similarly ,

$\displaystyle \frac{1}{f_2} = (\mu_2 – 1)( \frac{1}{-R} – \frac{1}{\infty} )$

$\displaystyle \frac{1}{f_2} = – \frac{\mu_2 – 1}{R} $

The equivalent focal length (f) of the given lens is

$\displaystyle \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} $

$\displaystyle \frac{1}{f} = \frac{\mu_1 – 1}{R} – \frac{\mu_2 – 1}{R} $

$\displaystyle \frac{1}{f} = \frac{\mu_1 – \mu_2}{R} $

$\displaystyle f = \frac{R}{\mu_1 – \mu_2} $