Physics - Geometrical Optics Question with Solution | TestHub

PhysicsGeometrical OpticsRefraction+ReflectionMedium2 minQB

PhysicsMediuminteger

Passage / Comprehension

In figure L is half part of an equiconvex lens of refractive index 1.5 whose surfaces have radius of curvature 40 cm and its right surface is silvered as shown in figure. Normal to its principal axis, a plane mirror M is placed on right of lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens. Images formed by lens and mirror coincide with each other and transverse length of final image formed by the lens is twice that of image formed by the mirror.

The value of b (in cm) is :-

Answer:

Solution:

The silvered half-lens acts as a concave mirror of power $P = 2 P_{L} + P_{M}$ :

$\frac{1}{f} = (1.5 - 1) \cdot \frac{2}{40} + \frac{2 \times 1.5}{40} = \frac{1}{40} + \frac{3}{40} \Rightarrow f = - 10 cm$ .

Lens image is twice the mirror image, so the lens magnification is 2: $\frac{a + 2 b}{a} = 2 \Rightarrow a = 2 b$ .

$\frac{1}{a + 2 b} - \frac{1}{a} = - \frac{1}{10} \Rightarrow - \frac{1}{2 a} = - \frac{1}{10}$ gives a = 5 cm, so b = 2.5 cm. Answer: 2.5.

Stream:JEESubject:PhysicsTopic:Geometrical OpticsSubtopic:Refraction+Reflection

⏱ 2mℹ️ Source: QB

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