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Geometrical Optics question

2016 · Q129
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Geometrical Optics question

2016 · Q129

NEETPhysicsGeometrical OpticsMCQ+4 / −1
Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror]

Column 1 Column 2
(A) m = - 2 (p) Convex mirror
(B) m = -12{1 \over 2}21​ (q) Concave mirror
(C) m = +2 (r) Real image
(D) m = +12{1 \over 2}21​ (s) Virtual image
  1. A
    A  →\to→  p   and   s;   B →\to→ q and r;   C →\to→ q and s;   D →\to→ q and r
  2. B
    A  →\to→  r   and   s;   B →\to→ q and s;   C →\to→ q and r;   D →\to→ p and s
  3. C
    A  →\to→  q  and  r;   B →\to→ q and r;   C →\to→ q and s;   D →\to→ p and s
  4. D
    A  →\to→  p   and   r;   B →\to→ p and s;   C →\to→ p and q;   D →\to→ r and s
View written solutionFree

Correct answer: C

Magnification in the mirror, m = −vu - {v \over u}−uv​

m = –2 ⇒\Rightarrow⇒ v = 2u

As v and u have same signs so the mirror is concave and image formed is real.

m=−12m = - {1 \over 2}m=−21​ ⇒\Rightarrow⇒ v = u2{u \over 2}2u​

⇒\Rightarrow⇒ Concave mirror and real image.

m = + 2 ⇒\Rightarrow⇒ v = –2u


As v and u have different signs but magnification is 2 so the mirror is concave and image formed is virtual.

m=+12m = + {1 \over 2}m=+21​ ⇒\Rightarrow⇒ v = −u2 - {u \over 2}−2u​

As v and u have different signs with magnification 12{1 \over 2}21​ so the mirror is convex and image formed is virtual.

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