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.**The table below shows object and image distances for four objects placed in front of mirrors. For which one is the image formed by a convex spherical mirror. [Positive and negative signs are used in accordance with standard sign convention

Solution

For (a) and (b), object and image are on the same side, so image is real as the object is real For (c), image is virtual but |m|>1. So concave mirror For (d), |m|less than 1 and image is also virtual, so in (d) image is formed in a convex mirror

For (a) and (b), object and image are on the same side, so image is real as the object is real For (c), image is virtual but |m|>1. So concave mirror For (d), |m|less than 1 and image is also virtual, so in (d) image is formed in a convex mirror

**Q2.**The refractive index of a prism is 2. The prism can have a maximum refracting angle of

Solution

Critical angle Î¸_c=sin^(-1)(1/Î¼) =sin^(-1)(1/2)=30° If A>2Î¸_c, the ray does not emerge from the prism. So, maximum angle can be 60°

Critical angle Î¸_c=sin^(-1)(1/Î¼) =sin^(-1)(1/2)=30° If A>2Î¸_c, the ray does not emerge from the prism. So, maximum angle can be 60°

**Q3.**A point source of light S is placed in front of a perfectly reflecting mirror as shown in the figure. ∑is a screen. The intensity at the centre of screen is found to be I If the mirror is removed, then the intensity at the center of screen would be

Solution

Let power of light source be P, then intensity at any point on the screen is due to light rays directly received from source and that due to light rays after reflection from the mirror I=P/(4Ï€a^2 )+P/(4Ï€×(3a)^2 )

When mirror is taken away,

I_1=P/(4Ï€a^2 )=9I/10

Let power of light source be P, then intensity at any point on the screen is due to light rays directly received from source and that due to light rays after reflection from the mirror I=P/(4Ï€a^2 )+P/(4Ï€×(3a)^2 )

When mirror is taken away,

I_1=P/(4Ï€a^2 )=9I/10

**Q4.**The refracting angle of a prism is

Solution

From sin((A+Î´_m)/2)=Î¼ sinA/2 sin((A+Î´_m)/2)=cosA/2/sinA/2 ×sinA/2

(A+Î´_m)/2=Ï€/2-A/2

Î´_m=Ï€-2A=180°-2A

From sin((A+Î´_m)/2)=Î¼ sinA/2 sin((A+Î´_m)/2)=cosA/2/sinA/2 ×sinA/2

(A+Î´_m)/2=Ï€/2-A/2

Î´_m=Ï€-2A=180°-2A

**Q5.**A point object is kept in front of a plane mirror. The plane mirror is doing SHM of amplitude 2 cm. The plane mirror moves along the

**Q6.**An eye specialist prescribes spectacles having a combination of convex lens of focal length

Solution

Power of convex lens P_1=100/40=2.5 D

Power of concave lens P_2=-100/25=-4 D

Now P=P_1+P_2=2.5 D-4 D=-1.5 D

Power of convex lens P_1=100/40=2.5 D

Power of concave lens P_2=-100/25=-4 D

Now P=P_1+P_2=2.5 D-4 D=-1.5 D

**Q7.**The lateral magnification of the lens with an object located at two different positions

Solution

For lens 1/f=1/v_1 -1/u_1 u_1/f=u_1/v_1 -1 ⇒ m_1=v_1/u_1 =f/(u_1+f)

And m_2=f/(u_2+f)

1/m_2 -1/m_1 =((u_2-u_1))/f ⇒ f=((u_2-u_1))/((m_2 )^(-1)-(m_1 )^(-1) )

For lens 1/f=1/v_1 -1/u_1 u_1/f=u_1/v_1 -1 ⇒ m_1=v_1/u_1 =f/(u_1+f)

And m_2=f/(u_2+f)

1/m_2 -1/m_1 =((u_2-u_1))/f ⇒ f=((u_2-u_1))/((m_2 )^(-1)-(m_1 )^(-1) )

**Q8.**In a thick glass slab of thickness

Solution

Shift =(l-m)(1-1/n_1 )+m(1-1/n_2 )=0

Shift =(l-m)(1-1/n_1 )+m(1-1/n_2 )=0

**Q9.**A ray of light enters a rectangular glass slab of refractive index

**Q10.**

A
convex lens is in contact with concave lens. The magnitude of the ratio of
their focal length is 2/3. Their equivalent focal length is 30 cm. What are
their individual focal lengths? |

Solution

Let focal length of convex lens is +f then focal length of concave lens would be -3/2 f. From the given condition,

1/30=1/f-2/3f=1/3f

∴ f=10 cm

Therefore, focal length of convex lens = + 10 cm and that of concave lens = -15 cm.

Let focal length of convex lens is +f then focal length of concave lens would be -3/2 f. From the given condition,

1/30=1/f-2/3f=1/3f

∴ f=10 cm

Therefore, focal length of convex lens = + 10 cm and that of concave lens = -15 cm.