Subjects like physics are very conceptual and the CBSE exams have numerical problems to test students’ ability to apply the concepts. It is very important that the student knows the correct approach to any problem while solving the same. If the student can approach the question correctly, most of the time his steps will be right and he will get the final answer without any problems. The following questions illustrate what needs to be kept in mind and how to approach the question in a planned manner.
A convex lens of focal length 20cm is placed 10cm in front of a convex mirror of radius 15cm. Where should a point object be placed in front of the lens so that it images on itself?
When a question such as this is asked, in most cases the student starts to find the image by applying lens formulae and mirror formulae for each optical element and this leads to mistakes. Whereas, in optics normally we find a system of optical elements with stipulation that the image is formed on the object itself. In such cases there has to be a mirror at the end of the optical system and rays have to be incident normally on the mirror in order to retrace the path.
While solving the problem, the student should keep in mind that when the image is formed on the object itself, then the ray must form normally on the mirror.
Converging rays are intercepted by a concave lens of focal length 20cm. If the rays originally converged to a point 10cm in front of the lens, where will they now converge after passing through the lens?
In the Rays and Optics chapter, this is an important question, in which most students get confused as to what the virtual object is.
In most cases the student assumes that, when the converging ray falls on an optical element, the object is kept at infinity. The student needs to understand and remember, like in the above case, whenever converging rays fall on an optical element, the object is virtual. In the absence of the optical element, the point of convergence is the position of virtual object and object distance from optical element to the point of convergence behind the optical element.
How does charge, capacity, potential difference, electric field and energy associated with a capacitor change qualitatively when separation between its plates is increased with battery (a) connected (b) disconnected?
The important part to keep in mind is the fact that when the battery remains connected, then the potential difference across the capacitor remains constant and when the battery is disconnected, then charge on the capacitor remains constant.
The student gets confused and makes the mistake by ignoring the above fact and starts to solve the question. Moreover, the battery connected implies that potential difference between the plates remains constant. Whereas battery disconnected implies that charge is held constant. Also capacitance of a capacitor depends only on geometry. So increasing separation between the plates decreases the capacitance.
(The authori is head, department of physics, HT Studymate)