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**Q.1. If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to **** (2020)****(1) 22 mm(2) 12 mm(3) 2 mm(4) 33 mmAns.** (1)

Given that m = 375, f

Magnification of compound microscope is given by

Where L is length of tube, D is near point of eye D = 25 cm, f

(1) 1

(2) 9

(3) 5

(4) 17

Ans.

Solution.

Given that,

According to Lens Makerâ€™s formula, we have

For air, we have

For liquid, we have

So,

(1) 15Â°

(2) 30Â°

(3) 45Â°

(4) 60Â°

Ans.

Solution.

Given that

Reflective index is given by

Critical angle is given by

Q.4. The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece? (2020)

Ans.

Let length of the tube, focal length of eye piece lens and focal length of objective lens be L, f

We know that

Given that L = 60cm, M = 5

So, f

f

From Eqs. (1) and (2), we get

f

Ans.

Solution.

Given that n = 1.5, R

From lens maker formula, we have

f = 60 cm

(1)

**(2)****(3)****(4)****Ans.** (3)**Solution.**

Lens formula is given by.............. (1)

Magnification is given by m = v/u ............. (2)

At u = f Eqs. (1) and (2) becomes

v = âˆž, m = âˆž

At u = 2f Eqs. (1) and (2) becomes

v = -2f, m = âˆ’1

So, only graph of option (3) satisfy the conditions.**Q.7 The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is 4 Ã— 10 ^{5} km. With light of wavelength of 5500 Ã…, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to (2020)(1) 60 m(2) 20 m(3) 200 m(4) 600 mAns. **(1)

Solution.

For just resolved the angular separation is given by

.................. (1)

Where D is aperture diameter of a telescope.

Let distance between object and surface of the moon be x and distance between earth and moon be d. So, Î¸ = x/d ............ (2)

From Eqs. (1) and (2), we get

Given that

Therefore,

(1)

(2)

(3)

(4)

Ans.

Solution.

Given condition is shown in the following figure

Apparent depth of the bottom of the vessel is given by

Apparent depth of the bottom between two liquids, that is, at height h from bottom is

Apparent depth of the bottom at the top of the vessel, that is, at height 2h from bottom is

(1) 20Â°

(2) 30Â°

(3) 25Â°

(4) 45Â°

Ans.

Solution.

Given that Î» = 6000 x 10

n

For 2

............. (1)

For 1

sinÎ¸

From Eq. (1), we get

=> Î¸ âˆ¼ 25Â°

(1) 71.6Â°

(2) 18.4Â°

(3) 90Â°

(4) 45Â°

Ans.

Solution.

Given that I = I

Where I

According to Malus Law, we have I = I

From Eqs. (1) and (2), we get

Î¸ = 71.57Â°

So, angle by which analyzer need to be rotated to reduce output intensity to zero is

Î± = 90 âˆ’Î¸ = 90 - 71.57 = 18.43Â°

(1) 6.9 mm

(2) 3.9 mm

(3) 5.9 mm

(4) 4.9 mm

Ans.

Solution.

Given that, Î» = 589 nm = 589 x 10

We know, separation between the successive bright fringes is given by x = Î»D/d

= 5.89 mm ~ 5.9 mm

(1) 0.853

(2) 0.672

(3) 0.568

(4) 0.760

Ans.

Solution.

Given that

So,

Intensities at phase difference is given by

= 0.853

(1) 21%

(2) 34%

(3) 17%

(4) 50%

Ans.

Let the angle of incident of light on surface of water be Î¸.

Given conditions is shown by following figure

Since, there is total internal reflection.

From Snellâ€™s law, we have n

Here

So,

Sin Î¸ = 3/4

And cos Î¸ = âˆš7/4

Area of sphere where light spread is given by Î© = 4Ï€r

The area of a spherical cap (at the surface) is given by

The percentage of light that emerges out of surface is given by

Ans.

Given that Î»

Distance of n

Where D is the distance of screen from slits and d is the distance between slits.

So,

Since, x

Therefore, n

15 x 500 = 10 x Î»

Î»

(1) 1.1 cm away from the lens.

(2) 0

(3) 0.55 cm towards the lens.

(4) 0.55 cm away from the lens.

Ans.

Distance of convex lens from screen, v = 10 cm

Refractive index, Î¼ = 1.5 Thickness of glass, t = 1.5 cm

Therefore, f = 5 cm

Now, the slab is shifted in the direction of incident ray

Distance of convex lens from source of light, u = âˆ’(10 â€“ 0.5) = â€“9.5 cm

Hence, the screen is shifted 0.55 cm away from the lens.

(4) 4/3

Ans.

According to Brewesterâ€™s law,

Putting this value in Eq. (1), we get

sin i

Squaring both the sides, we get

Thus, the minimum value of Î¼ is

(1) 45Â°

(2) 60Â°

(3) 75Â°

(4) 90Â°

Ans.

As we know that, sum of angle of triangle = 180Â°

Î¸ + Î¸ + Î¸ = 180Â°

â‡’ 3Î¸ = 180Â°

â‡’ Î¸ = 60Â°

(1) Î¼

(2) 2Î¼

(3) 3Î¼

(4) 2Î¼

Ans.

We know that

Put these values in Eq. (1), we get

(1) 1 cm

(2) 2 cm

(3) 4.0 cm

(4) 3.1 cm

Ans.

It is given that

(1) 1.16 Ã— 10

(2) 3.22 Ã— 10

(3) 0.92 Ã— 10

(4) 2.26 Ã— 10

Ans.

Therefore, v

= (0.015)

= 1.15 Ã— 10

D

Since, on increasing the wavelength, Î¼ decreases and hence D

(1) 90Â°

(2) 30Â°

(3) 60Â°

(4) 45Â°

Ans.

(1) 70 cm from point B at right; virtual.

(2) 40 cm from point B at right; real.

(3) 20/3 cm from point B at right, real.

(4) 70 cm from point B at right; real.

Ans.

For second lens refraction,

Therefore, 70 cm from B at right, real.

(1) d

(2) 2d

(3) 3d

(4) d/2

Ans.

Consider Î”ICD and Î”IAB.

Since, both are similar triangle so, ratio of identical side of triangle are equal.

(1) f

(4) f

Ans.

For plano-concave lens

When both lenses are combined:

(1) Image disappears

(2) Magnified image

(3) Erect real image

(4) No change

Ans.

Solution. Focal length

Since, whole set up is immersed in water without disturbing the object and screen position. Thus, focal length of lens will change hence image disappear from the screen.

(1) 16 : 9

(2) 25 : 9

(3) 4 : 1

(4) 5 : 3

Ans.

Solution. The ratio of the maximum intensity to the minimum intensity

We know that,

where, A

Using componendo and dividendo rule, we have

Thus, the required ratio is 25 : 9.

(1) 640

(2) 320

(3) 321

(4) 641

Ans.

Path difference

dsinÎ¸ = nÎ»

0.32 Ã— 10

Therefore, n = 320

(1) 625 nm, 500 nm

(2) 380 nm, 525 nm

(3) 380 nm, 500 nm

(4) 400 nm, 500 nm

Ans.

Ans.

At point P we have

For first minima we know that,

(1) 0.94

(2) 0.80

(3) 0.74

(4) 0.85

Ans.

(1) 10

(2) 05

(3) 04

(4) 09

Ans.

Second minima is

For interference, path difference at P

Path difference at Q

So, order of maxima is 5, 6, 7, 8, 9

Therefore, 5 bright fringes are there between the first and second diffraction minima.

(1) 30 V/m

(2) 10 V/m

(3) 24 V/m

(4) 6 V/m

Ans.

I' = 0.96I

Incident energy in air

Incident energy in glass

Put these values in Eq. (1), we get

**(2) **

**(3) **

**(4) Ans:** (3)

(1) 0

(2) 30

(3) 45

(4) 60

Ans:

Unpolarized light of intensity I, when passed through a polarizer A, its intensity becomes I/2

Since intensity of light emerging from polarizer B = I/2

So, A & B are parallel placed.

Let, C makes angle Î¸ with A.

So,

(i.e. distance between the centres of each slit.) [2018]

(1) 25 Î¼m

(2) 50 Î¼m

(3) 75 Î¼m

(4) 100 Î¼m

Ans:

(1) Ï€/8

(2) Ï€/12

(3) Ï€/6

(4) Ï€/4

Ans:

b sin Î¸ = nÎ»

b sin Î¸ = 2Î»

sin Î¸ = 1/2

Î¸ = Ï€/6

(1)

Ans:

Wave propagation vector should be along

So, 2 is the only option which satisfies the above condition.

**Q 39. A particle is oscillating on the X-axis with an amplitude 2 cm about the point x _{0} = 10 cm, with a frequency Ï‰. A concave mirror of focal length 5 cm is placed at the origin (see figure). [2018]Identify the correct statements.**

(B) The image executes non-periodic motion.

(C) The turning points of the image are asymmetric w.r.t. the image of the point at x =10 cm.

(D) The distance between the turning points of the oscillation of the image is 100/21 cm.

(1) (A), (D)

(2) (A), (C), (D)

(3) (B), (C)

(4) (B), (D)

Ans:

For mean

As image copies the time period of object (A) is right as well. It will be periodic motion

For one extreme

Right arrow = -40/3 cm

for other extreme

These points are asymmetric about x

Amplitude of oscillation of image

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