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Q1
A convex lens of focal length 1.0 m and a concave lens of focal length 0.25 m are apart. A parallel beam of light is incident on the convex lens. The beam emerging after refraction from both lenses is
Q2
A convex lens A of focal length 20 cm and a concave lens G of focal length 5 cm are kept along the same axis with the distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then distance d in cm will be
Q3
A convex lens forms an image of an object placed 20 cm away from it at a distance of 20 cm on the other side of the lens. If the object is moves 5 cm toward the lens, the image will be
Q4
With a concave mirror, an object is placed at a distance x1 from the principal focus, on the principal axis. The image is formed at a distance x2from the principal focus. The focal length of the mirror is
Q5
A concave mirror is placed on a horizontal table with its axis direction vertically upward. Let O be the pole of the mirror and C its center of curvature. A point object is placed at C. It has a real image, also located atC. If the mirror is now filled with water, the image will be
Q6
A cubical room is formed with six plane mirrors. An insect moves along the diagonal of the floor with uniform speed. The velocities of its image in two adjacent walls are
Q7
A plane glass mirror of thickness 3 cm of material of
Q8
A point source of light B is placed at a distance L is front of the center of a mirror of width d hung vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown in figure, the greatest distance over which he can see the image of the light source in the mirror is
Q9
An object is placed at a distance of 25 cm from the pole of a convex mirror and a plane mirror is set at a distance 5 cm from convex mirror so that the virtual images formed by the two mirrors do not have any parallax. The focal length of the convex mirror is
Q10
A convex mirror of radius of curvature 1.6 m has an object placed at a distance of 1 m from it. The image is formed at a distance of
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`10 ms^(-1)``1 ms^(-1)``9 ms^(-1)``0.9 ms^(-1)`
Answer : B
Solution : Mirror formula, `1/u + 1/v = 1/f` <br> Differentiating with respect to t, we get <br> `:. - (1)/(u^(2)) (du)/(dt) - (1)/(v^(2)) (dv)/(dt) = 0`, ( ` :'` f is contant) <br> or `(dv)/(dt) = - (v/u)^(2) (du)/(dt)` <br> `v_(i) = -(v/u)^(2)v_(o)` , ( `:' (dv)/(dt)= v_(i)` and `(du)/(dt) = v_(o)`) <br> Substituting the given values, we get <br> `v_(i) = -(10/30)^(2) xx 9 = - 1 mx^(-1)` or `|v_(i)| = 1 ms^(-1)`