Physics

$v1 −u1 =f1 $

or, $v1 =301 −401 $

or, $v1 =1204−3 $

or, $v=+120cm$

Now, this image acts as the object for the lens.

So, $u=+120cm,f=30cm$

$v1 −u1 =f1 $

or, $v1 =301 +1201 $

or, $v=+24cm$

Two cases are possible.

Case 1:

Final image is formed by the concave lens.

For concave lens $v=∞$; $f=−20cm$

$⇒$ $u=+20cm$

Now, $v=20+15=35cm$ serves as image distance for the convex lens.

So, $v=+35cm$; $f=+30cm$

$v1 −u1 =f1 $

$⇒$ $351 −u1 =301 $

$⇒$ $u=−210cm=210cm$ from converging lens.

Case 2:

Final image is formed by convex lens.

So,$v=∞$; $f=+30cm$

$⇒$ $u=−30cm$ for convex lens to form image at infinity

So for the concave lens, the image distance is $v=−30−(−15)=−15cm$; $f=−20cm$

$v1 −u1 =f1 $

$⇒$ $−151 −u1 =−201 $

or, $u=−60cm$, i.e, $60cm$ from diverging lens.

$v1 −u1 =f1 $

or, $v1 −−301 =20−1 $

or, $v=60cm$

So, $d=3×h=0.1×3=0.3cm$

$m=uv =h_{o}h_{i} $$=3060 =2=0.05h_{i} $

or, $h_{i}=0.1cm$

Solution:

When the ray from O passes through the slab of refractive index ($μ$), then there will be shift of point O to $I_{1}$ and then this point $I_{1}$ will act as source for the concave mirror.

Shift = $(1−μ1 )t$$=(1−32 )6$

$⟹$ shift $=2cm$, i.e., the object will appear to look closer by $2cm$.

Now as the final image is formed at a point O itself, so the ray from point $I_{1}$ will retrace its own path (i-e, $I_{1}$ should be at $R$ of concave mirror).

So, $d−2+12=2×f_{1}=2×18=36$

$⟹d=26cm$

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