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. In the above question, the magnification is
Solution
m=-v/u=-(4/9)/(-1)=4/9
m=-v/u=-(4/9)/(-1)=4/9
Q2.A glass prism has refractive index √2and refracting angle 30°. One of the refracting surface of the prism is silvered. A beam of monochromatic light will retrace it path if its angle of incidence on the unsilvered refracting surface of the prism is
Q3. A point object is placed at a distance of 25 cm from a convex lens of focal length 20 cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image if formed at infinity. The thickness t is
Solution
Image will be formed at infinity if object is placed at focus of the lens, i.e., at 20 cm from the lens. Hence, Shift =25-20=(1-1/μ)μ Or 5(1-1/1.5)tor t=(5×1.5)/0.5=15 cm
Image will be formed at infinity if object is placed at focus of the lens, i.e., at 20 cm from the lens. Hence, Shift =25-20=(1-1/μ)μ Or 5(1-1/1.5)tor t=(5×1.5)/0.5=15 cm
Q4. A lens forms a virtual, diminished image of an object placed at 2 m from it. The size of image is half of the object. Which one of the following statements is correct regarding the nature and focal length of the lens?
Solution
m=f/(f+u),1/2=f/(f-2); or 2f=f-2 or f=-2 m |f|=2 metre. Since the image is virtual as well as diminished, therefore the lens is concave
m=f/(f+u),1/2=f/(f-2); or 2f=f-2 or f=-2 m |f|=2 metre. Since the image is virtual as well as diminished, therefore the lens is concave
Q5.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
Solution
Clearly, 2f=20 cm Or f=10 cm Now, u=-15 cm, v=? f=10 cm 1/v-1/(-15)=1/10 Or 1/v+1/15=1/10 or1/v=1/10-1/15 Or 1/v=(3-2)/30=1/30 Or v=30 cm The change in image distance is (30-20) cm, i.e., 10 cm
Clearly, 2f=20 cm Or f=10 cm Now, u=-15 cm, v=? f=10 cm 1/v-1/(-15)=1/10 Or 1/v+1/15=1/10 or1/v=1/10-1/15 Or 1/v=(3-2)/30=1/30 Or v=30 cm The change in image distance is (30-20) cm, i.e., 10 cm
Q7. A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true?
Solution
During minimum deviation the ray inside the prism is parallel to the base of the prism in case of an equilateral prism.
During minimum deviation the ray inside the prism is parallel to the base of the prism in case of an equilateral prism.
Q8. A plane mirror is placed at origin parallel to y-axis, facing the positive x-axis. An object starts from (2 m, 0, 0) with a velocity of (2i ̂+2j ̂ ) ms^(-1). The relative velocity of the image with respect to the object is along
Q9.A transparent sphere of radius 20 cm and refractive index 1.6 is fixed in a hole of the partition separating the two media: A (refractive index n_1=1.2) and B(refractive index n_3=1.7). A luminous point object is placed 120 cm from the surface of the sphere in medium A. It is viewed from D in medium B in a direction normal to the sphere. Find the position of the image formed by the rays, from point N
Q10. A point object is placed in front of a thick plane mirror as shown in figure below. Find the location of final image w.r.t. object