## A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite position….

Q: A small object stuck on the surface of a glass sphere (n = 1.5) is viewed from the diametrically opposite position. Find transverse magnification.

Click to See Solution :
Ans: m = 3
Sol:

$\displaystyle \frac{n_2}{v} – \frac{n_1}{u} = \frac{n_2 – n_1}{R}$

$\displaystyle \frac{1}{v} – \frac{1.5}{-2R} = \frac{1 – 1.5}{-R}$

v = -4 R

$\displaystyle m = \frac{n_1 v}{n_2 u }$

$\displaystyle m = \frac{1.5 (-4R)}{1 (-2R) }$

m = 3

## A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 ….

Q: A parallel beam of light is incident on a lens of focal length 10 cm. A parallel slab of refractive index 1.5 and thickness 3 cm is placed on the other side of the lens. Find the distance of the final image from the lens.

Click to See Solution :
Sol: As rays are parallel to the principal axis, image is created by lens at the focus. By placing of glass-slab,
Shift $\displaystyle = (1-\frac{1}{\mu})t$

$\displaystyle = (1-\frac{1}{1.5}) 3= 1 cm$

Irrespective of separation,
Image is shifted to the right by 1 cm. Total distance from lens 10 + 1 = 11 cm

## In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column …

Q: In Resonance tube experiment, if 400 Hz tuning fork is used, the first resonance occurs when length of air column in the tube is 19 cm. If the 400 Hz. tuning fork is replaced by 1600 Hz tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correction = 1 cm)

(A) 5 cm

(B) 10 cm

(C) 15 cm

(D) 20 cm

Click to See Solution :
Ans: (A),(C)

Sol: For first resonance with 400 Hz tuning fork

$\displaystyle L_{eq} = \frac{v}{4 f_0 }$

$\displaystyle L_{eq} = \frac{v}{4 (400) } = (19+1) = 20 cm$

$\displaystyle \frac{v}{4 f_0 } = \frac{v}{4 (1600) } = \frac{20}{4} = 5 cm$

For Resonance

$\displaystyle L_{eq} = \frac{v}{4 f_0 } , \frac{3 v}{4 f_0 } , \frac{5 v}{4 f_0 }, \frac{7 v}{4 f_0 }….$

1 cm + L = 5 cm , 15 cm , 25 cm , 35 cm , 45 cm …..

L = 4 cm , 14 cm , 24 cm , 34 cm , 44 cm …..

water level should be further lowered by
24 – 19 = 5 cm

⇒ 34 – 19 = 15 cm

## A man is holding an umbrella at angle 30° with vertical with lower end towards himself, which is appropriate angle ….

Q: A man is holding an umbrella at angle 30° with vertical with lower end towards himself, which is appropriate angle to protect him from rain for his horizontal velocity 10 m/s. Then which of the following will be true-

(A) rain is falling at angle 30° with vertical, towards the man

(B) rain may be falling at angle 30° with vertical, away from the man

(C) rain is falling vertically

(D) none of these

Click to See Solution :
Ans: (B)
Sol:

Let v = velocity of rain ; Hence Possible values of α are
-30º < α < 90º