question_answer

(i) Use the mirror equation to show that a convex mirror always produces a virtual image independent of the location of the object. An object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

(ii) Use the lens equation deduce algebraically that an object placed with in the focus of a convex lens produces a virtual and enlarged image.

Answer:

(i) For convex mirror, \[\text{f}>0\] Also,     \[u<0\] But from mirror equation,             \[\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}=\frac{1}{\text{v}}+\frac{1}{(-\,\text{u})}\] [putting u with sign] \[\frac{1}{\text{v}}=\frac{1}{\text{f}}+\frac{1}{\text{u}}\]             \[\Rightarrow \]   If f and u to be positive, then                                     \[\frac{1}{\text{v}}>0\,\,\,\,\,\,\Rightarrow \,\,\,\,\text{v0}\]             \[\because \] Linear magnification, \[m=\frac{-\,\text{v}}{\text{u}}\] \[\Rightarrow \]   \[m=\frac{-\,\text{v}}{-\,\text{u}}\]        [putting with sign]                         \[=\frac{\text{v}}{\,\text{u}}\] \[\therefore \]      \[m>0\] Hence, the virtual image is formed by the convex mirror. For concave mirror,\[\text{f}\,\,\text{}\,\,\text{0,}\] \[\text{u}\,\,\text{}\,\,\text{0,}\] \[\left| \text{f} \right|>\left| \text{u} \right|>0\]                                                             [given] But from mirror equation,                         \[\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}\] \[\Rightarrow \]   \[\frac{-1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}\]         [putting with sign] \[\frac{1}{\text{v}}=\frac{1}{\text{u}}-\frac{1}{\text{f}}\Rightarrow \frac{1}{\text{v}}=\frac{\text{f}-\text{u}}{\text{fu}}\] \[\because \]       \[\left| \text{f} \right|>\left| \text{u} \right|\]                                \[\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{\text{u}-\text{f}}{\text{uf}}\]      (numerically) \[\Rightarrow \]   \[\frac{1}{\left| \text{v} \right|}>0\Rightarrow \text{v}>0\] Image is formed on RHS of mirror. Also,                 \[\frac{1}{\text{f}}=\frac{1}{\left| \text{v} \right|}-\frac{1}{\left| \text{u} \right|}\] For concave mirror, Linear magnification, \[m=\frac{-\,\text{v}}{\text{u}}\] \[=\frac{-\,\text{v}}{-\,\text{u}}\]                       [putting with sign] \[=\frac{\text{v}}{\,\text{u}}\] \[\therefore \]      \[m>1\] Enlarged virtual and erected image is formed on the other side of mirror. (ii) For convex lens, \[\text{f}>0\] Also,     \[\text{u}>0\] But from lens formula, \[\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}\]             \[=\frac{1}{\text{v}}+\frac{1}{\text{u}}\]  [taking u with negative sign] \[\Rightarrow \]   \[\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{\text{u}-\text{f}}{\text{uf}}\]            (numerically) Since, given       \[\text{u}\,\,\text{f}\]                  (numerically) Thus,    \[\frac{1}{\text{v}}<0\Rightarrow \text{v}<0\] Image is formed on LHS of the lens i.e., virtual image. For convex lens, f is positive. \[\Rightarrow \]   \[\frac{1}{\left| \text{v} \right|}<\frac{1}{\left| \text{u} \right|}\Rightarrow \frac{\left| \text{v} \right|}{\left| \text{u} \right|}>1\Rightarrow m>1\] Enlarged virtual image formed on LHS of lens.