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# Integer Value of Ray and Wave Optics, Past year Questions JEE Advance, Class 12, Physics JEE Notes | EduRev

## JEE : Integer Value of Ray and Wave Optics, Past year Questions JEE Advance, Class 12, Physics JEE Notes | EduRev

The document Integer Value of Ray and Wave Optics, Past year Questions JEE Advance, Class 12, Physics JEE Notes | EduRev is a part of the JEE Course Class 12 Physics 35 Years JEE Mains &Advance Past year Paper.
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Q.1. The focal length of a thin biconvex lens is 20 cm. When an object is moved from a distance of 25 cm in front of it to 50 cm, the magnification of its image changes from m25 to m50. The ratio         (2010)

Ans. 6

Solution.

Q.2. A large glass slab (Î¼= 5 / 3) of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius R cm. What is the value of R?     (2010)

Ans. 6

Solution. In the figure, C represents the critical angle

Q.3. Image of an object approaching a convex mirror of radius of curvature 20 m along its optical axis is observed to move from 25/3 m to 50/7 m in 30 seconds. What is the speed of the object in km per hour?  (2010)

Ans. 3

Solution. Using mirror formula for first position

Using mirror formula for the second position

Change in position of object = 25 m

Q.4. Water (with refractive index = 4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature â€˜R = 6 cmâ€™ as shown. Consider oil to act as a thin lens. An object â€˜Sâ€™ is placed 24 cm above water surface. The location of its image is at â€˜xâ€™ cm above the bottom of the tank. Then â€˜xâ€™ is     (2011)

Ans. 2

Solution. For the convex spherical refracting surface of oil we apply

âˆ´ v = 21 cm

For water-oil interface

âˆ´ V' = 16 cm.

This is the image distance from water-oil interface.

Therefore the distance  of the image from the bottom of the tank is 2 cm.

Q.5. A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maximum on the surface of water are given by x2 = p2m2Î»2 â€“ d2, where l is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is      (JEE Adv. 2015)

Ans. 3

Solution. For maxima

Path defference = mÎ»

âˆ´ S2A â€“ S1A = mÎ»

Q.6. Consider a concave mirror and a convex lens (refractive index = 1.5) of focal length 10 cm each, separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M1. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M2. The magnitude  is         (JEE Adv. 2015)

Ans. 7

Solution. Applying mirror formula

For convex lens u = |2fâ„“

Therefore image will have a magnification of 1.

When the set â€“ up is kept in a medium

The focal length of the lens will change

â‡’  = 17.5 cm.

Applying lens formula

= Magnification by lens

Q.7. The monochromatic beam of light is incident at 60Â° on one face of an equilateral prism of refractive index n and emerges from the opposite face making an angle Î¸(n) with the normal (see the figure). For n = âˆš3 the value of q is 60Â° and

The value of m is                     (JEE Adv. 2015)

Ans. 2

Solution. Here âˆ MPQ + âˆ MQP  = 60Â°. If âˆ MPQ = r then âˆ MQP

= 60 â€“ r

Applying Snellâ€™s law at P

sin60Â° = n sin r                    ...(i)

Differentiating w.r.t â€˜nâ€™ we get

...(ii)

Applying Snellâ€™s law at Q

sin Î¸ = n sin (60Â° â€“ r) ...(iii)

Differentiating the above equation w.r.t â€˜nâ€™ we get

[from (ii)]

[sin (60Â° â€“ r) + cos (60Â° â€“ r) tan r] ...(iv)

From eq. (i), substituting n = âˆš3 we get r = 30Â°

From eq (iii), substituting n = âˆš3 , r = 30Â° we get Î¸ = 60Â°

On substituting the values of r and Î¸ in eq (iv) we get

[sin 30Â° + cos 30Â° tan 30Â°] = 2

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