JEE Main 2019 (Online) 8th April Morning Slot | Ray & Wave Optics Question 176 | Physics

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be :

20 cm from the convergent mirror, same size as the object

40 cm from the convergent mirror, same size as the object

40 cm from the convergent lens, same size as the object

20 cm from the convergent mirror, twice the size of the object

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

In figure, the optical fiber is $$l$$ = 2m long and has a diameter of d = 20 μm. If a ray of light is incident on one end of the fiber at angle $$\theta _1$$ = 40°, the number of reflection it makes before emerging from the other end is close to: (refractive index of fibre is 1.31 and sin 40° = 0.64) JEE Main 2019 (Online) 8th April Morning Slot Physics - Ray & Wave Optics Question 177 English

JEE Main 2019 (Online) 8th April Morning Slot Physics - Ray & Wave Optics Question 177 English

57000

55000

66000

45000

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

-1

In an interference experiment the ratio of amplitudes of coherent waves is $${{{a_1}} \over {{a_2}}} = {1 \over 3}$$ . The ratio of maximum and minimum intensities of fringes will be :

JEE Main 2019 (Online) 12th January Evening Slot

MCQ (Single Correct Answer)

-1

A plano-convex lens (focal length f₂, refractive index $$\mu $$₂, radius of curvature R) fits exactly into a plano-concave lens (focal length f₁, refractive index $$\mu $$₁, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :

f₁ + f₂

f₁ $$-$$ f₂

$${R \over {{\mu _2} - {\mu _1}}}$$

$${{2{f_1}{f_2}} \over {{f_1} + {f_2}}}$$

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