A symmetric biconvex lens of the radius of curvature R and made of glass of refractive index 1 . 5 , is placed on a layer of liquid placed on top of a plane mirror as shown in the figure. An optical needle with its tip on the principal axis of the lens is moved along the axis until its real, inverted image coincides with the needle itself. The distance of the needle from the lens is measured to be x . On removing the liquid layer and repeating the experiment, the distance is found to be y . Obtain the expression for the refractive index of the liquid in terms of x and y .

We are given, the refractive index of the lens, μ g = 1 . 5 . The distance of the needle from the lens in the first case is equal to the focal length of the combination of a convex lens and a planoconcave lens formed by the liquid. Thus, f = x The distance measured in the second case is equal to the focal length of the convex lens. Thus, f 1 = y Let f 2 be the focal length of the planoconcave lens formed by the liquid. Then, 1 f = 1 f 1 + 1 f 2 1 f 2 = 1 x - 1 y = y - x x y f 2 = x y y - x Let the radius of curvature of the first surface be R The radius of curvature of the second surface will be - R Thus, for the biconvex lens, 1 f 1 = μ g - 1 1 R - 1 - R 1 y = 1 . 5 - 1 2 R ⇒ y = R Let the refractive index of liquid be μ . For plano-convex lens, R 2 = ∞ R 1 = - R Thus, 1 f 2 = μ - 1 1 - R - 1 ∞ Substituting values from above, y - x x y = μ - 1 1 - y μ - 1 = - ( y - x ) x μ = x - y x + 1 μ = 2 x - y x

State Gauss’ law for magnetism. Explain its significance.

Gauss law for magnetism says that if a closed surface is imagined in a magnetic field, the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero. Gauss law signifies that magnetic mono poles does not exist.Every closed surface has magnetic flux in the inward direction to the south pole and equal flux in the outward direction to the north pole for a dipole. Since magnetic flux is not zero in the case of mono poles, it is a fact that mono poles do not exist.

E M B I B E

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise-Choice>Q 22

HARD

12th Karnataka Board

IMPORTANT

Earn 100

State and obtain Malus law.

Important Questions on Solved Physics Paper-2

EASY

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise-Choice>Q 23

Find De-Broglie wavelength of

150 eV

electron.

HARD

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise-Choice>Q 24

State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits ?

MEDIUM

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise-Choice>Q 25

Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.

EASY

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise>Q 26

If electric potential at a point is zero then electric field intensity must be zero.Give reasons with an example.

EASY

12th Karnataka Board

IMPORTANT

Physics Crash Course (Based on Revised Syllabus-2023)>Chapter 16 - Solved Physics Paper-2>Exercise>Q 27

Twelve resistors each of resistance $r$ are joined to form a cube as shown in the figure given below. Find the equivalent resistance between the diagonally opposite points $A$ and $F$ .

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