## RAY OPTICS AND OPTICAL INSTRUMENTS - 6

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

Q1.Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is
•  30° for both the colours
•  Greater for the violet colour
•  Greater for the red colour
•  Equal but not 30° for both the colours
Solution
At minimum deviation (Î´=Î´_m ) r_1=r_2=A/2=(60°)/2=30° (for both colours)
Q2.Refraction takes place at a concave spherical boundary separating glass-air medium. For the image to be real, the object distance (Î¼_g=3/2)
•  Should be greater than three times the radius of curvature of the refracting surface
•  Should be greater than two times the radius of curvature of the refracting surface
•  Should be greater than the radius of curvature of the refracting surface
•  Is independent of the radius of curvature of the refracting surface
Solution

Q3.  A fish looks upward at an unobstructed overcast sky. What total angle does the sky appear to subtend? (Take refractive index of water as √2)
•   180°
•  90°
•  75°
•  60°
Solution
The fish can observe the sky only if refraction takes place. If TIR takes place, then image of sky cannot be observed i.e., i is less than i_c and i_c=45° So, angle subtended =Ï€/2

Q4. A particle revolves in clockwise direction (as seen from point A) in a circle C of radius 1 cm and completes one revolution in 2 sec. The axis of the circle and the principal axis of the mirror M coincide, call it AB. The radius of curvature of the mirror is 20 cm. Then, the direction of revolution (as seen from A) of the image of the particle and its speed is
•  Î¼>Clockwise, 1.57 cms^(-1)
•  Clockwise, 3.14 cms^(-1)
•  Î¼>Anticlockwise, 1.57 cms^(-1)
•  Anticlockwise, 3.14 cms^(-1)
Solution

By mirror formula :1/v+1/(-10)=1/10 ⇒v=+5 cm ∴m=+1/2 The image revolves in a circle of radius 1/2 cm. Image of a radius is erect ⇒ particle will revolve in the same direction as the particle. The image will complete one revolution in the same time 2 s Velocity of image v is Ï‰r=2Ï€/2×1/2=Ï€/2 cms^(-1)=1.57 cms^(-1)
Q5. A cube of side 2 m is placed in front of a concave mirror of focal length 1 m with its face A at a distance of 3 m and face B at a distance of 5 m form the mirror. The distance between the images of faces A and B and heights of images of A and B are, respectively,
•  1 m, 0.5 m, 0.25 m
•  0.5 m, 1 m, 0.25 m
•  0.5 m, 0.25 m, 1 m
•  0.25 m, 1 m, 0.5 m
Solution

Q6. The focal lengths of the objective and the eye-piece of a compound microscope are 2.0 cm and 3.0 cm respectively. The distance between the objective and the eye-piece is 15.0 cm. The final image formed by the eye-piece is at infinity. The two lenses are thin. The distances in cm of the object and the image produced by the objective measured from the objective lens are respectively
•  2.4 and 12.0
•  2.4 and 15.0
• 2.3 and 12.0
•  2.3 and 3.0
Solution

When final image is formed at infinity, length of the tube =v_o+f_e ⇒15=v_o+3⇒v_o=12 cm For objective lens 1/f_o =1/v_o -1/u_o ⇒1/((+2))=1/((+12))-1/u⇒u_o=-2.4 cm
Q7.The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the lens is nearly
•  3 diopter
•  5 diopter
•  2 diopter
•  9 diopter
Solution

Q8.A convex lens of focal length 20 cm and a concave lens of focal length f are mounted coaxially 5 cm apart. Parallel beam of light incident on the convex lens emerges from the concave lens as a parallel beam. Then, f in cm is
•  35
•  T25
•  20
•  15
Solution

Clearly, power of system is zero ∴0=1/20+1/f-5/20f Or -1/20=15/20f or f=-15 cm

Q9.A right-angled prism of apex angle 4° and refractive index 1.5 is located in front of a vertical plane mirror as shown in figure. A horizontal ray of light is falling on the prism. Find the total deviation produced in the light ray as it emerges 2nd time from the prism
•  8° CW
•  6° CW
•  180° CW
•  174° CW
Solution

Deviation produced by prism is Î´_1=(Î¼-1)A=2° CW Angle of incidence for mirror is Î´_1, so deviation produced by the mirror is Î´_2=Ï€-2Î´_1=176° CW
Deviation produced by the prism for 2nd refraction is Î´_3=2° ACW Net deviation =174° CW
Q10. A U-shaped wire is placed before a concave mirror having radius of curvature 20 cm as shown in figure. Find the total length of the image
•  2 cm
•  10 cm
•  8 cm
• 14 cm
Solution

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