Plane glass slab is kept between a model and a plane mirror. The mirror is 50 cm away from the model. The glass slab has a thickness of 15 cm and refractive index of 1.5. At what distance (in cm ) from herself will the model see her image?

Question

Plane glass slab is kept between a model and a plane mirror. The mirror is 50 cm away from the model. The glass slab has a thickness of 15 cm and refractive index of 1.5. At what distance (in cm ) from herself will the model see her image? (A) 40 (B) 50 (C) 80 (D) 90

Tardigrade Admin · Accepted Answer

Δ S =15(1-(1/1.5))=(15/3)=5 cm Mirror is at 50-5=45 cm ⇒ Image is 45 cm behind the mirror. Final image =45+50-5=90 cm

Q. Plane glass slab is kept between a model and a plane mirror. The mirror is $50 c m$ away from the model. The glass slab has a thickness of $15 c m$ and refractive index of $1.5.$ At what distance (in $c m$ ) from herself will the model see her image?

40

50

80

90

$Δ S = 15 (1 - \frac{1}{1.5}) = \frac{15}{3} = 5 c m$
Mirror is at $50 - 5 = 45 c m$
$\Rightarrow$ Image is $45 c m$ behind the mirror.
Final image $= 45 + 50 - 5 = 90 c m$

Q. Plane glass slab is kept between a model and a plane mirror. The mirror is 50cm away from the model. The glass slab has a thickness of 15cm and refractive index of 1.5. At what distance (in cm ) from herself will the model see her image?

40

50

80

90

ΔS=15(1−1.51​)=315​=5cm Mirror is at 50−5=45cm ⇒ Image is 45cm behind the mirror. Final image =45+50−5=90cm

Q. Plane glass slab is kept between a model and a plane mirror. The mirror is $50 c m$ away from the model. The glass slab has a thickness of $15 c m$ and refractive index of $1.5.$ At what distance (in $c m$ ) from herself will the model see her image?

$Δ S = 15 (1 - \frac{1}{1.5}) = \frac{15}{3} = 5 c m$
Mirror is at $50 - 5 = 45 c m$
$\Rightarrow$ Image is $45 c m$ behind the mirror.
Final image $= 45 + 50 - 5 = 90 c m$