WB JEE 2021 | Geometrical Optics Question 9 | Physics

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

A spherical convex surface of power 5 D separates object and image space of refractive indices 1.0 and $$4\over3$$ , respectively. The radius of curvature of the surface is

20 cm

1 cm

4 cm

5 cm

WB JEE 2021

MCQ (Single Correct Answer)

-0.25

WB JEE 2021 Physics - Geometrical Optics Question 9 English

The cross-section of a reflecting surface is represented by the equation x² + y² = R² as shown in the figure. A ray travelling in the positive x-direction is directed toward positive y-direction after reflection from the surface at point M. The coordinate of the point M on the reflecting surface is

$$\left( {{R \over {\sqrt 2 }},{R \over {\sqrt 2 }}} \right)$$

$$\left( { - {R \over 2}, - {R \over 2}} \right)$$

$$\left( { - {R \over {\sqrt 2 }},{R \over {\sqrt 2 }}} \right)$$

$$\left( {{R \over {\sqrt 2 }}, - {R \over {\sqrt 2 }}} \right)$$

WB JEE 2020

MCQ (Single Correct Answer)

-0.25

A tennis ball hits the floor with a speed v at an angle $$\theta $$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $$\varepsilon $$, what will be the angle of reflection?

$${\tan ^{ - 1}}\left( {{{\tan \theta } \over \varepsilon }} \right)$$

$${\sin ^{ - 1}}\left( {{{\sin \theta } \over \varepsilon }} \right)$$

$$\theta $$$$\varepsilon $$

$$\theta {{2\varepsilon } \over {\varepsilon + 1}}$$

WB JEE 2020

MCQ (Single Correct Answer)

-0.25

An object, is placed 60 cm in front of a convex mirror of focal length 30 cm. A plane mirror is now placed facing the object in between the object and the convex mirror such that it covers lower half of the convex mirror. What should be the distance of the plane mirror from the object, so that there will be no parallax between the images formed by the two mirrors?

40 cm

30 cm

20 cm

15 cm

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